qu.1.topic=Ch 5, Sec 1 - Newton's Second Law of Motion@ qu.1.1.mode=Randomized Formula@ qu.1.1.editing=useHTML@ qu.1.1.name=Decelerating Car@ qu.1.1.question=You are traveling in a car at $v mi/h. At some point you see an obstacle ahead and start braking until the car comes to a stop. Find the decelerating force exerted on the car by the brakes if the mass of the car is $m tons and your stopping distance is $d m. (Assume the deceleration is constant and there is no deceleration due to friction).@ qu.1.1.answer=$ans N (1 ? 0.05)@ qu.1.1.random= $v\=rand(50,75,3); $m\=rand(0.8,1.5,3); $d\=rand(45,70,3); $ans/=sig(3,$m*1000*($v*1.6/3.6)^2/(2*$d))@ qu.1.1.hint.1=Newton's second law of motion states that an object with mass m has an acceleration a equal to the net force ΣF acting on that object divided by its mass m: a = ΣF / m.@ qu.1.1.comment= First you must solve for the acceleration of the car using the equation for constant acceleration. The problem gives v₀, v, and x, so you can use the following equation to solve for a: v² = v₀² + 2ax. Once you have acceleration, simply substitute the appropriate values into the equation ΣF = ma. Note that mass m, must be in kg.@ qu.1.2.mode=Randomized Formula@ qu.1.2.editing=useHTML@ qu.1.2.name=Get up to Speed@ qu.1.2.question=A constant force of $f N acts on a $m-kg $body. If the $body starts from rest, how much time will it take the $body to get up to speed of $v m/s?@ qu.1.2.answer=$ans s (1 ? 0.05)@ qu.1.2.random= $f\=rand(30,300,3); $m\=rand(10,50,3); $body\=switch(rint(2),"body","object"); $v\=rand(2,15,3); $ans\=sig(3,$m*$v/$f)@ qu.1.2.hint.1=Newton's second law of motion states that an object with mass m has an acceleration a equal to the net force ΣF acting on that object divided by its mass m: a = ΣF / m.@ qu.1.2.comment= First you must solve for the acceleration of the $body using Newton's second law of motion a = ΣF / m. Then, you know a, v₀, and v, so you can use the following equation to solve for t: v = v₀ + at.@ qu.1.3.mode=Randomized Formula@ qu.1.3.editing=useHTML@ qu.1.3.name=Zero Acceleration@ qu.1.3.question=A particle is moving with zero acceleration while forces F₁ and F₂ are acting on it. If F₁ = ($i N) i + ($j N) j, what must be the magnitude of F₂?@ qu.1.3.answer=$ans N (1 ? 0.05)@ qu.1.3.random= $i\=rand(5,50,3); $j\=rand(10,30,3); $ans\=sig(3,(($i)^2+($j)^2)^0.5)@ qu.1.3.hint.1=Newton's second law of motion states that an object with mass m has an acceleration a equal to the net force ΣF acting on that object divided by its mass m: a = ΣF / m.@ qu.1.3.hint.2=If acceleration is zero, consider what ΣF must be in order to make the equation ΣF = ma true.@ qu.1.3.comment= 1. If acceleration is zero, consider what ΣF must be in order to make the equation ΣF = ma true.
2. If the acceleration of a particle is zero, the net force acting on that particle must also be zero; thus, the forces F₁ and F₂ must sum to zero. So, the magnitude of F₂ must be equal to that of F₁, but they must be pointing in opposite directions. Thus, use the Pythagorean theorem to find the magnitude of F₁ from its components, and this must also be the magnitude of F₂.@ qu.1.4.mode=Randomized Formula@ qu.1.4.editing=useHTML@ qu.1.4.name=Free Particle@ qu.1.4.question=A free particle that has a mass of $m × 10^{- $mpower} kg is initially at rest. At some point it is accelerated by a force. After the particle travels $d m, it has a speed of $v × 10^$power m/s. What is the magnitude of the accelerating force?@ qu.1.4.answer=$ans N (1 ? 0.05)@ qu.1.4.random= $mpower1\=switch(rint(9),23,24,25,26,27,28,29,30,31); $mpower\=int($mpower1); $m\=rand(1.1,9.9,3); $mass\=$m*10^(-$mpower); $d\=rand(15,50,3); $v\=rand(1.1,9.9,3); $power1\=switch(rint(6),2,3,4,5,6,7); $power\=int($power1); $speed\=$v*10^$power; $ans\=sig(3,$mass*($speed)^2/(2*$d))@ qu.1.4.hint.1=Newton's second law of motion states that an object with mass m has an acceleration a equal to the net force ΣF acting on that object divided by its mass m: a = ΣF / m.@ qu.1.4.comment= In order to calculate the accelerating force, you must first find the acceleration of the particle. The problem gives values of v₀, x, and v, so you can use the following equation to solve for acceleration, a: v² = v₀² + 2ax. Once you know the acceleration, you can substitute the appropriate values into the equation ΣF = ma to find the magnitude of the accelerating force.@ qu.1.5.mode=Randomized Formula@ qu.1.5.editing=useHTML@ qu.1.5.name=Plow Through Snow@ qu.1.5.question=A driver in a $m-kg $vehicle attempting to get through a snowdrift approaches it at a speed of $v mi / h. The $vehicle plows through $d m of snow before coming to a complete stop. What is the magnitude of the force exerted on the $vehicle by the snow?@ qu.1.5.answer=$ans N (1 ? 0.05)@ qu.1.5.random= $vehicle\=switch(rint(3),"vehicle","car","truck"); $m\=rand(900,1500,3); $v\=rand(35,55,3); $d\=rand(10,50,3); $ans\=sig(3,$m*($v*1.6/3.6)^2/(2*$d))@ qu.1.5.hint.1=Newton's second law of motion states that an object with mass m has an acceleration a equal to the net force ΣF acting on that object divided by its mass m: a = ΣF / m.@ qu.1.5.comment=In order to calculate the force exerted by the snow, you must first obtain deceleration of the vehicle. The problem gives values for v, v₀, and x, so you can use the following equation to solve for acceleration, a: v² = v₀² + 2ax. Once you know the acceleration, you can substitute the appropriate values into the equation ΣF = ma to find the magnitude of the force exerted by the snow.@ qu.1.6.mode=Randomized Formula@ qu.1.6.editing=useHTML@ qu.1.6.name=Shopping Cart@ qu.1.6.question=A $m-kg shopping cart is moving with constant velocity in the absence of friction. At some point in time it reaches a $deg° incline. Find the decelerating force that acts on the cart while it is moving up the incline.@ qu.1.6.answer=$ans N (1 ? 0.05)@ qu.1.6.random= $m\=rand(10,25,3); $deg\=rand(5,25,3); $ans\=sig(3,$m*9.8*sin($deg*3.142/180))@ qu.1.6.hint.1=The only forces acting on the shopping cart are gravitational force and the normal force (the force exerted by the ground on the cart). Remember that only forces applied in the direction opposite to the direction of motion cause deceleration.@ qu.1.6.comment= 1. The only forces acting on the shopping cart are gravitational force and the normal force (the force exerted by the ground on the cart). Remember that only forces applied in the direction opposite to the direction of motion cause deceleration.

2. If you place the x-axis of the coordinate system along the incline, you will see that the only force causing deceleration of the cart (the only force pointing opposite to the direction of motion) is the horizontal component of the gravitational force. Using trigonometric functions, you can solve for the magnitude of this force: F = mg×sin θ.@ qu.1.7.mode=Randomized Formula@ qu.1.7.editing=useHTML@ qu.1.7.name=Shopping Cart 1@ qu.1.7.question=A shopping cart is moving with constant velocity of $v m/s in the absence of friction. At some point in time it reaches a $deg° incline. Counting from the bottom of the incline, what distance does the cart travel up the incline before it momentarily comes to a stop?@ qu.1.7.answer=$ans m (1 ? 0.05)@ qu.1.7.random= $deg\=rand(5,25,3); $v\=rand(3,7,3); $ans\=sig(3,(($v)^2)/(2*9.8*sin($deg*3.142/180)))@ qu.1.7.hint.1=Newton's second law of motion states that an object with mass m has an acceleration a equal to the net force ΣF acting on that object divided by its mass m: a = ΣF / m.@ qu.1.7.hint.2=The only forces acting on the shopping cart are gravitational force and the normal force (the force exerted by the ground on the cart). Remember that only forces applied in the direction opposite to the direction of motion cause deceleration.@ qu.1.7.hint.3=First you must know the magnitude of the decelerating force in order to find the magnitude of deceleration. This will then allow you to find the distance the cart will travel before it momentarily comes to a stop.@ qu.1.7.comment= 1. The only forces acting on the shopping cart are gravitational force and the normal force (the force exerted by the ground on the cart). Remember that only forces applied in the direction opposite to the direction of motion cause deceleration.
2. First you must know the magnitude of the decelerating force in order to find the magnitude of deceleration. This will then allow you to find the distance the cart will travel before it momentarily comes to a stop.
3. If you place the x-axis of the coordinate system along the incline, you will see that the only force causing deceleration of the cart (the only force pointing opposite to the direction of motion) is the horizontal component of the gravitational force. Using trigonometric functions, you can find the magnitude of the decelerating force to be equal to mg×sin θ. Then you can write down an expression for Newton’s Second Law as follows: mg×sin θ = ma, which yields to a = g×sin(θ). Use this value of acceleration in the equation v² = v₀² + 2ax to solve for the distance the cart will travel.@ qu.1.8.mode=Randomized Formula@ qu.1.8.editing=useHTML@ qu.1.8.name=Trailer@ qu.1.8.question=A truck accelerates a $m-kg trailer from rest up to a speed of $v mi/h over a distance of $d m. What is the force exerted on the truck by the trailer? Assume that the acceleration of the trailer is constant.@ qu.1.8.answer=$ans N (1 ? 0.05)@ qu.1.8.random= $m\=rand(500,3000,3); $v\=rand(45,65,3); $d\=rand(100,300,3); $ans\=sig(3,$m*($v*1.6/3.6)^2/(2*$d))@ qu.1.8.hint.1=Newton's second law of motion states that an object with mass m has an acceleration a equal to the net force ΣF acting on that object divided by its mass m: a = ΣF / m.@ qu.1.8.hint.2=By Newton's third law, force exerted by the truck on the trailer is equal to the force that the trailer exerts on the truck.@ qu.1.8.comment= 1. By Newton's third law, force exerted by the truck on the trailer is equal to the force that the trailer exerts on the truck.
2. In order to solve for the force exerted on the truck by the trailer, you must first know the magnitude of acceleration. Since the problem gives values for v₀, v, and x, you can use the following equation to solve for acceleration a: v² = v₀² + 2ax. After you obtain a, you can substitute the appropriate values into the equation ΣF = ma to solve for the force.@ qu.1.9.mode=Randomized Formula@ qu.1.9.editing=useHTML@ qu.1.9.name=Trailer 1@ qu.1.9.question=A truck begins to pull a trailer from rest accelerating in up to $v mi/h over $d m. What is the mass of the trailer if the force exerted by the truck on the trailer is $f N?@ qu.1.9.answer=$ans kg (1 ? 0.05)@ qu.1.9.random= $v\=rand(40,50,3); $d\=rand(200,300,3); $f\=rand(5000,7000,3); $a\=($v*1397/3125)^2/(2*$d); $ans\=sig(3,$f/$a)@ qu.1.9.hint.1=Newton's second law of motion states that an object with mass m has an acceleration a equal to the net force ΣF acting on that object divided by its mass m: a = ΣF / m.@ qu.1.9.comment= In order to calculate the mass of the trailer, you must first determine the magnitude of acceleration of the trailer. Since the problem gives values for v₀, v, and x, you can use the following equation to solve for acceleration a: v² = v₀² + 2ax. After you obtain a, you can substitute the appropriate values into the equation ΣF = ma to solve for the mass of the trailer.@ qu.2.topic=Ch 5, Sec 2 - Newton's Third Law of Motion@ qu.2.1.mode=Randomized Formula@ qu.2.1.editing=useHTML@ qu.2.1.name=Wall@ qu.2.1.question=$object1 is thrown against a wall. When it hits the wall, the $object exerts a force of $f N on the wall. Calculate the force exerted on the $object by the wall.@ qu.2.1.answer=$f N@ qu.2.1.random= $integer\=rint(3); $object1\=switch($integer,"An object","A ball","A body"); $object\=switch($integer,"object","ball","body"); $f\=rand(20,90,3)@ qu.2.1.hint.1=Newton’s third law of motion states that forces come in pairs. Those forces are equal in magnitude but opposite in direction, or, more commonly, “for every action there is an equal and opposite reaction.”@ qu.2.1.comment=By Newton’s third law, the force applied by the ball on the wall is equal in magnitude to the force the wall applies on the ball, though the directions of these forces are different.@ qu.2.2.mode=Randomized Formula@ qu.2.2.editing=useHTML@ qu.2.2.name=Three Boxes@ qu.2.2.question=Three boxes are placed side by side. A worker pushes the boxes with the force of $f N directed parallel to the ground. $qu if the masses of the boxes A, B, and C are equal $ma kg, $mb kg, and $mc kg respectively.

@ qu.2.2.answer=$ans N (1 ? 0.05)@ qu.2.2.random= $integer\=rint(2); $f\=rand(7,30,3); $qu\=switch($integer,"Find the force that the box B exerts on box A","Find the force that the box C exerts on box B"); $ma\=rand(2,4,3); $kb\=rand(0.1,0.4,3); $kc\=rand(0.25,0.6,3); $mb\=sig(3,$ma-$ma*$kb); $mc\=sig(3,$ma+$ma*$kc); $a\=$f/($ma+$mb+$mc); $calculation\=switch($integer,$ma*$a,$mb*$a); $ans\=sig(3,$calculation)@ qu.2.2.hint.1=Newton’s third law of motion states that forces come in pairs. Those forces are equal in magnitude but opposite in direction or, more commonly, “for every action there is an equal and opposite reaction”.@ qu.2.2.hint.2=To find the acceleration of one of the boxes (their common acceleration), treat all three boxes as one large object and use their combined mass and the applied force to solve for the acceleration.@ qu.2.2.comment= 1. To find the acceleration of one of the boxes (the common acceleration), treat all three boxes as one large object and use their combined mass and the applied force to solve for the acceleration.
2. Use the common acceleration of the boxes and the mass of the box, on which the force is being exerted, in the ΣF = ma equation to find the force exerted on that box.@ qu.2.3.mode=Randomized Formula@ qu.2.3.editing=useHTML@ qu.2.3.name=Driver in Car@ qu.2.3.question=A $m-kg person driving a car accelerates up a $angle° incline at $a m/s². Find the magnitude of the force exerted on the driver by the car.@ qu.2.3.answer=$ans N (1 ? 0.05)@ qu.2.3.random= $angle\=rand(7,27,3); $a\=rand(1.7,3.5,3); $m\=rand(60,90,3); $calculation\=$m*$a; $ans\=sig(3,$calculation)@ qu.2.3.hint.1=If you place the coordinate system on the incline (with the x axis along the incline), note that there is no acceleration in the y direction, so the normal force (the force exerted by the surface on the car) and the y component of gravity are equal resulting in no net force in the y direction.@ qu.2.3.hint.2=The force exerted on the driver by the car equals to the net force acting on the driver. According to Newton’s second law of motion, what is the net force acting on a body equal to?@ qu.2.3.comment=According to Newton’s second law of motion, the net force acting on a body equals to the product of the mass of the body and its acceleration (ΣF = ma).@ qu.2.4.mode=Randomized Formula@ qu.2.4.editing=useHTML@ qu.2.4.name=Elevator@ qu.2.4.question=A $m-kg $person is riding in an elevator that is accelerating $downward at $a m/s². What is the magnitude of the force exerted on the elevator by the $person?@ qu.2.4.answer=$ans N (1 ? 0.05)@ qu.2.4.random= $integer\=rint(2); $person\=switch(rint(3),"scientist","physicist","person"); $m\=rand(60,90,3); $downward\=switch($integer,"downward","upward"); $a\=rand(0.8,3,3); $calculation\=switch($integer,$m*9.8-$m*$a,$m*9.8+$m*$a); $ans\=sig(3,$calculation)@ qu.2.4.hint.1=The force exerted on the elevator by the $person is equal in magnitude but opposite to the force exerted on the $person by the elevator. The force exerted on the $person by the elevator is also called the normal force.@ qu.2.4.hint.2=Suppose we direct the x axis upward. If the elevator is accelerating downward, the accelerating force is pointing opposite to the normal force, reducing the net force in the x direction. If the elevator is accelerating upward, the accelerating force is pointing in the same direction as the normal force, increasing the net force in x direction.@ qu.2.4.comment=The force exerted on the $person by the elevator can be found using Newton’s second law of motion.
Keeping in mind that the axis is directed upward, we can write down the equations projected on that axis for two cases.
Case of the elevator accelerating upward: ma = - mg + N.
Case of the elevator accelerating downward: - ma = - mg + N.
Finally, the magnitude of the force exerted on the elevator by the person is equal to the magnitude of N.@ qu.2.5.mode=Randomized Formula@ qu.2.5.editing=useHTML@ qu.2.5.name=Crash Test@ qu.2.5.question=On a crash test, a $m-kg $vehicle moving at $v mi/h is directed toward a high-density concrete wall. What is the magnitude of the force exerted by the $vehicle on the wall if the impact took place over $t s?@ qu.2.5.answer=$ans N (1 ? 0.05)@ qu.2.5.random= $vehicle\=switch(rint(4),"vehicle","car","truck","mini van"); $m\=rand(900,1500,3); $v\=rand(45,65,3); $t\=rand(0.07,0.21,3); $a\=($v*1.6/3.6)/$t; $ans\=sig(3,$m*$a)@ qu.2.5.hint.1=Newton’s third law of motion states that forces come in pairs. Those forces are equal in magnitude but opposite in direction or, more commonly, “for every action there is an equal and opposite reaction.”@ qu.2.5.hint.2=In order to find the force exerted by the $vehicle on the wall, we need to find the force exerted on the $vehicle by the wall, which is the stopping force applied to the $vehicle. Use Newton’s second law of motion to find this force.@ qu.2.5.comment= According to Newton’s second law, the magnitude of the stopping force acting on the $vehicle is expressed by the following equation: F_stopping = ma (assuming there is no friction).
Finally, the magnitude of the force exerted by the vehicle on the wall equals to F_stopping. @ qu.3.topic=Ch 5, Sec 3 - Normal Forces@ qu.3.1.mode=Multiple Choice@ qu.3.1.editing=useHTML@ qu.3.1.name=Pulled@ qu.3.1.question=A $block is being pulled by a force F as shown below. What is the possible direction of the normal force vector N?

@ qu.3.1.answer=3@ qu.3.1.random= $block\=switch(rint(2),"brick","block")@ qu.3.1.choice.1=

@ qu.3.1.choice.2=

@ qu.3.1.choice.3=

@ qu.3.1.choice.4=

@ qu.3.1.choice.5=

@ qu.3.1.comment=A normal force is defined as a force exerted by a surface on a body. That force is perpendicular to the surface.@ qu.3.2.mode=Multiple Choice@ qu.3.2.editing=useHTML@ qu.3.2.name=Pushed Against Wall@ qu.3.2.question=A $block is pushed against a wall by a force F as shown below. Assuming that the $block is held in place, what is the direction of the normal force N?

@ qu.3.2.answer=5@ qu.3.2.choice.1=

@ qu.3.2.choice.2=

@ qu.3.2.choice.3=

@ qu.3.2.choice.4=

@ qu.3.2.choice.5=

@ qu.3.2.random= $block\=switch(rint(2),"brick","block")@ qu.3.2.comment=A normal force is defined as a force exerted by a surface on a body. That force is perpendicular to the surface.@ qu.3.3.mode=Multiple Choice@ qu.3.3.editing=useHTML@ qu.3.3.name=On Incline@ qu.3.3.question=Consider a block at rest on an incline. Which of the following pictures illustrates the correct direction of the normal force N?@ qu.3.3.answer=1@ qu.3.3.choice.1=

@ qu.3.3.choice.2=

@ qu.3.3.choice.3=

@ qu.3.3.choice.4=

@ qu.3.3.choice.5=

@ qu.3.3.comment=A normal force is defined as a force exerted by a surface on a body. That force is perpendicular to the surface.@ qu.4.topic=Ch 5, Sec 4 - Free-Body Diagrams@ qu.4.1.mode=Randomized Formula@ qu.4.1.editing=useHTML@ qu.4.1.name=Pulled Box@ qu.4.1.question=An overhead view of a $m-kg box being pulled by three people is shown below. Find the magnitude of acceleration of the box if F₁ = $f1 N, F₂ = $f2 N, F₃ = $f3 N, and if the angles a, b, and g equal $alpha°, $beta°, and $gamma° respectively.

@ qu.4.1.answer=$ans m/s^2 (1 ? 0.05)@ qu.4.1.random= $m\=rand(20,70,3); $f1\=rand(12,27,3); $f2\=rand(4,25,3); $f3\=rand(10,50,3); $alpha\=rand(10,50,3); $beta\=rand(20,70,3); $gamma\=rand(10,80,3); $force_y\=($f1)*sin($alpha*3.142/180)+($f2)*sin($beta*3.142/180)-($f3)*cos($gamma*3.142/180); $force_x\=($f2)*cos($beta*3.142/180)-($f1)*cos($alpha*3.142/180)+($f3)*sin($gamma*3.142/180); $a_x\=$force_x/$m; $a_y\=$force_y/$m; $ans\=sig(3,(($a_x)^2+($a_y)^2)^(1/2))@ qu.4.1.hint.1= Approach this problem by applying Newton’s second law of motion to the box. Using the data given in the text of the problem, we can easily find the net force acting on the box in each direction. Because the box accelerates in two directions (x and y directions), you will have two equations that contain a_x and a_y respectively.@ qu.4.1.comment= 1. Approach this problem by applying Newton’s second law of motion to the box. Using the data given in the text of the problem, we can easily find the net force acting on the box in each direction. Because the box accelerates in two directions (x and y directions), you will have two equations that contain a_x and a_y respectively.
2. After you solve for a_x and a_y, the magnitude of acceleration can be found by using Pythagorean Theorem.@ qu.4.2.mode=Randomized Formula@ qu.4.2.editing=useHTML@ qu.4.2.name=Carriage & Horses@ qu.4.2.question=Below is the diagram of a carriage pulled by two horses. Find the magnitude of the force that each of the horses exerts on the carriage. The mass of the carriage is $mass kg, the acceleration is $a m/s², and a = $alpha°. Assume that there is no friction and that both horses pull equally hard.

@ qu.4.2.answer=$ans N (1 ? 0.05)@ qu.4.2.random= $mass\=rand(350,500,3); $a\=rand(0.8,3,3); $alpha\=rand(15,40,3); $ans\=sig(3,$mass*$a/(2*cos($alpha*3.142/180)))@ qu.4.2.hint.1= Approach this problem by applying Newton’s second law of motion to the carriage. Using the data given in the text of the problem, we can find the net force acting on the carriage in each direction. Because the horses pull equally hard, the y components of the forces they exert will cancel each other out. Thus, the acceleration in the y direction is equal to zero (a_y = 0 m/s²).@ qu.4.2.comment= 1. Approach this problem by applying Newton’s second law of motion to the carriage. Using the data given in the text of the problem, we can find the net force acting on the carriage in each direction. Because the horses pull equally hard, the y components of the forces they exert will cancel each other out. Thus, the acceleration in the y direction is equal to zero (a_y = 0 m/s²).
2. Finally, for the magnitude of the force F we get: F = ma_x / (2cosα).@ qu.4.3.mode=Randomized Formula@ qu.4.3.editing=useHTML@ qu.4.3.name=Cart@ qu.4.3.question=A worker pushes a $m-kg cart up an incline with a force of $f N directed horizontally. Find the $qu if a = $alpha°. Assume there is no friction.

@ qu.4.3.answer=$ans $units (1 ? 0.05)@ qu.4.3.random= $integer\=rint(2); $qu\=switch($integer,"normal force acting on the cart","acceleration of the cart"); $m\=rand(20,70,3); $coefficient\=rand(0.5,.9,3); $alpha\=rand(10,40,3); $f1\=(1+$coefficient)*$m*9.81*sin($alpha*3.142/180); $f\=sig(3,$f1); $units\=switch($integer,"N","m/s^2"); $calculation\=switch($integer,$m*9.8*cos($alpha*3.142/180)+$f*sin($alpha*3.142/180), ($f*cos($alpha*3.142/180)-$m*9.8*sin($alpha*3.142/180))/$m); $ans\=sig(3,$calculation)@ qu.4.3.hint.1=

According to the way we positioned the coordinate system (see picture), the cart does not accelerate in the y direction. Write down the equations for Newton's second law along each of the axes. Then, having two equations and two unknowns, you will be able to solve for the desired quantity.
2. Newton's second law for the cart along the two axes is written as follows.
Along x: F (cos α) – mg (sin α) = ma_x
Along y: N – mg (cos α) – F (sin α) = 0
Solve the system of equations for the desired quantity.@ qu.4.4.mode=Randomized Formula@ qu.4.4.editing=useHTML@ qu.4.4.name=File Cabinet@ qu.4.4.question=Two people are pushing a file cabinet as shown in the picture below. Find the magnitude of acceleration of the cabinet if F₁ = $f1 N, F₂ = $f2 N, a = $alpha°, b = $beta°, and the mass of the cabinet is $m kg. Assume there is no friction.

@ qu.4.4.answer=$ans m/s^2 (1 ? 0.05)@ qu.4.4.random= $k\=rand(0.1,0.4); $f1\=rand(7,25,3); $f2_\=$f1*(1-$k); $f2\=sig(3,$f2_); $m\=rand(50,100,3); $alpha\=rand(5,15,3); $beta\=sig(3,$alpha(1+$k)); $ans\=sig(3,($f1*cos($alpha*3.142/180)+$f2*cos($beta*3.142/180))/$m)@ qu.4.4.hint.1=Because the file cabinet accelerates in both x and y directions, we first need to find a_x and a_y (the x and y components of acceleration) and then use Pythagorean theorem to find the magnitude of total acceleration.
Write down the equations for Newton's second law along each of the axes.@ qu.4.4.comment= 1. Because the file cabinet accelerates in both x and y directions, we first need to find a_x and a_y (the x and y components of acceleration) and then use Pythagorean theorem to find the magnitude of total acceleration.
Write down the equations for Newton's second law along each of the axes.
2. According to Newton’s second law,
- ma_x = - F₁ cos α - F₂ cos β
and
- ma_y = - F₁ sin α - F₂ sin β.
Find a using Pythagorean Theorem (a² = a_x² + a_y²).@ qu.5.topic=Ch 5, Sec 5 - Other Applications of Newton's Laws@ qu.5.1.mode=Randomized Formula@ qu.5.1.editing=useHTML@ qu.5.1.name=Object in Elevator@ qu.5.1.question=An $object placed on a scale is riding in an elevator. What is the measurement of mass taken by an observer in the elevator if the mass of the $object at rest is $m kg? Assume that the acceleration of the elevator is directed $upward and is equal to $a m/s².@ qu.5.1.answer=$ans kg (1 ? 0.05)@ qu.5.1.random= $object\=switch(rint(2),"object","block"); $integer\=rint(2); $m\=rand(10,50,3); $a\=rand(0.7,3,3); $upward\=switch($integer,"upward","downward"); $calculation\=switch($integer,$m*(1+$a/9.81),$m*(1-$a/9.81)); $ans\=sig(3,$calculation)@ qu.5.1.hint.1= The reading of the scale will be equal to the force exerted on the scale by the object divided by g, the acceleration due to gravity. According to Newton’s third law, that force is equal in magnitude to the normal force acting on the object. We can find the normal force by applying Newton’s second law of motion to the object.@ qu.5.1.hint.2= Suppose we direct the x axis upward. If the elevator is accelerating downward, the accelerating force acting on the object is pointing opposite to the normal force, reducing the net force in the x direction. If the elevator is accelerating upward, the accelerating force acting on the object is pointing in the same direction as the normal force, increasing the net force in x direction.@ qu.5.1.comment= 1. The reading of the scale will be equal to the force exerted on the scale by the object divided by g, the acceleration due to gravity. According to Newton’s third law, that force is equal in magnitude to the normal force acting on the object. We can find the normal force by applying Newton’s second law of motion to the object.
2. Suppose we direct the x axis upward.
-- If the elevator is accelerating downward, the accelerating force acting on the object is pointing opposite to the normal force, reducing the net force in the x direction.
-- If the elevator is accelerating upward, the accelerating force acting on the object is pointing in the same direction as the normal force, increasing the net force in x direction.
3. The force exerted on the object by the scale can be found using Newton’s second law of motion. Keeping in mind that the axis is directed upward, we can write down the equations along that axis for two cases.
Case of the elevator accelerating upward: ma = - mg + N.
Case of the elevator accelerating downward: - ma = - mg + N.
Finally, the magnitude of the normal force acting on the object is equal to the magnitude of N.@ qu.5.2.mode=Randomized Formula@ qu.5.2.editing=useHTML@ qu.5.2.name=Sliding@ qu.5.2.question=A smooth $object slides down a $alpha° incline. What distance does the $object travel in $t s if it starts at rest? Assume there is no friction.@ qu.5.2.answer=$ans m (1 ? 0.05)@ qu.5.2.random= $object\=switch(rint(2),"body","object"); $t\=rand(0.5,5,3); $alpha\=rand(7,43,3); $ans\=sig(3,0.5*9.8*sin($alpha*3.142/180)*$t^2)@ qu.5.2.hint.1= Assume that the x axis is directed along the incline in the direction of motion of the $object. Write down the expression for Newton's second law of motion with respect to that axis and find the acceleration of the $object.@ qu.5.2.comment= 1. Assume that the x axis is directed along the incline in the direction of motion of the $object. Write down the expression for Newton's second law of motion with respect to that axis and find the acceleration of the $object.
2. Having found the acceleration of the $object, use the equation x = (¹/₂)at² to find the distance that the $object travels.@ qu.5.3.mode=Randomized Formula@ qu.5.3.editing=useHTML@ qu.5.3.name=Robot@ qu.5.3.question=A robot $picks a heavy $rock giving it an acceleration of $a m/s² and exerting a force of $f N. What is the weight of the $rock?@ qu.5.3.answer=$ans N (1 ? 0.05)@ qu.5.3.random= $picks\=switch(rint(2),"picks up","lifts"); $rock\=switch(rint(2),"rock","load"); $a\=rand(0.5,2,3); $f\=rand(700,3000,3); $m\=$f/(9.8+$a); $ans\=sig(3,$m*9.8)@ qu.5.3.hint.1= Write down Newton's second law for the $rock and find the mass of the load.@ qu.5.3.comment= 1. Write down Newton's second law for the $rock and find the mass of the load.
2. To find weight, multiply the mass of the $rock by g, which is the acceleration due to gravity.@