qu.1.topic=Ch 6, Sec 1 - General Questions About Friction@ qu.1.1.mode=Multiple Choice@ qu.1.1.editing=useHTML@ qu.1.1.name=Caused By@ qu.1.1.question=Friction is caused by __________.@ qu.1.1.answer=1@ qu.1.1.choice.1=irregularities of surfaces@ qu.1.1.choice.2=gravity@ qu.1.1.choice.3=applied forces@ qu.1.1.choice.4=chemical properties of materials@ qu.1.2.mode=Multiple Choice@ qu.1.2.editing=useHTML@ qu.1.2.name=fk Proportionality@ qu.1.2.question=Kinetic frictional force is proportional to the __________ of an object.@ qu.1.2.answer=5@ qu.1.2.choice.1=speed@ qu.1.2.choice.2=density@ qu.1.2.choice.3=weight@ qu.1.2.choice.4=mass@ qu.1.2.choice.5=none of these@ qu.1.2.comment= The equation for kinetic friction is as follows: F kinetic = μkN. This shows that kinetic friction is proportional to the normal force. It is true that the normal force can sometimes be equal in magnitude to the weight (mg) but this is not always the case. Thus, it is not completely correct to say that kinetic friction is proportional to weight.@ qu.1.3.mode=Multiple Choice@ qu.1.3.editing=useHTML@ qu.1.3.name=Static Friction (Body at Rest)@ qu.1.3.question=If a body is at rest, the static frictional force is __________.@ qu.1.3.answer=5@ qu.1.3.choice.1=maximum@ qu.1.3.choice.2=increasing@ qu.1.3.choice.3=decreasing@ qu.1.3.choice.4=zero@ qu.1.3.choice.5=the answer cannot be determined without additional information@ qu.1.3.comment= The static frictional force is a very adaptable force, meaning that as you apply more force the static force increases. Consider a very heavy box, if you push lightly on the box, it does not move, if you push slightly harder on the box, it still does not move. In both of these cases, there is no acceleration, so the static frictional force must be equal to the applied force in both cases. So, in this question, you cannot determine what is occurring in reference to the static frictional force without knowing what forces are being applied.@ qu.1.4.mode=Multiple Choice@ qu.1.4.editing=useHTML@ qu.1.4.name=Kinetic Friction -- Vector@ qu.1.4.question=The kinetic frictional force vector points in the direction __________.@ qu.1.4.answer=2@ qu.1.4.choice.1=of motion@ qu.1.4.choice.2=opposite to the direction of motion@ qu.1.4.choice.3=perpendicular to the direction of motion@ qu.1.4.choice.4=of the normal force exerted on a body@ qu.1.4.choice.5=the static frictional force vector would point if a body were at rest@ qu.1.5.mode=Multiple Choice@ qu.1.5.editing=useHTML@ qu.1.5.name=Static Friction -- Vector@ qu.1.5.question=Which of the following statements about the direction of the static frictional force vector are true?

I. It points in the direction of motion
II. It points in the direction opposite to the direction of motion
III. I does not have a direction
IV. It points in the direction opposite to the direction of attempted motion
V. It points in the direction of attempted motion
VI. It is sometimes directed perpendicular to the direction of motion
@ qu.1.5.answer=4@ qu.1.5.choice.1=I and II@ qu.1.5.choice.2=III only@ qu.1.5.choice.3=IV only@ qu.1.5.choice.4=IV and VI@ qu.1.5.choice.5=VI only@ qu.1.5.choice.5=V and VI@ qu.1.5.comment= Kinetic frictional force is directed opposite to the direction of motion, and the same is true for static friction – it is directed opposite to the direction of attempted motion. But there is the special case of circular motion, such as when a car is rounding a corner, when friction acts at a 90° angle to the direction of motion. This occurs, because the centrifugal force pushes the car sideways (away from center of the circle) and friction acts in a direction opposite to that force.@ qu.2.topic=Ch 6, Sec 2 - Kinetic Frictional Force@ qu.2.1.mode=Randomized Formula@ qu.2.1.editing=useHTML@ qu.2.1.name=Block Pushed Up a Wall@ qu.2.1.question=A $mass-kg block is pushed up a wall by a force F = $force N as shown in the picture. Find the magnitude of the acceleration of the block if the coefficient of kinetic friction between the wall and the block is $coeff_kinetic and q = $angle°.

@ qu.2.1.answer=$ans2 m/s^2 (1 ? 0.05)@ qu.2.1.random= $angle\=rand(10,37,3); $mass\=rand(1,5,3); $force\=rand(30,80,3); $coeff_kinetic\=rand(0.3,0.7,3); $ans\=sig(3,($force*cos($angle*3.14/180)-$coeff_kinetic*$force*sin($angle*3.14/180)-$mass*9.8)/$mass); $ans2\=abs($ans)@ qu.2.1.hint.1=To approach the problem, first draw a free body diagram and label all the forces acting on the block. Make sure to include a coordinate system and mark which direction is positive. Then apply Newton’s second law of motion in the x and y directions.@ qu.2.1.hint.2= ΣFx = N - F(sin θ) = 0
ΣFy = F(cos θ) – mg – μkN = ma@ qu.2.1.comment= 1. To approach the problem, first draw a free body diagram and label all the forces acting on the block. Make sure to include a coordinate system and mark which direction is positive. Then apply Newton’s second law of motion in the x and y directions.

2. ΣFx = N - F(sin θ) = 0
ΣFy = F(cos θ) – mg – μkN = ma

3. To solve the system of equations, find FN from the first equation and substitute this value in the second equation. Divide both sides of the second equation by m to find a.@ qu.2.2.mode=Randomized Formula@ qu.2.2.editing=useHTML@ qu.2.2.name=Block - Incline@ qu.2.2.question=A block (initially at rest at point A) starts to slide down a frictionless incline (refer to the picture below). When it reaches the rough horizontal floor, it decelerates and, eventually, comes to a complete stop at point B. How long does it take the block to travel from point A to point B if a = $angle°, d = $d m, and the coefficient of kinetic friction between the block and the rough horizontal floor is $coeff_kinetic?

@ qu.2.2.answer=$ans s (1 ? 0.05)@ qu.2.2.random= $angle\=rand(20,50,3); $coeff_kinetic\=rand(0.3,0.7,3); $d\=rand(1.5,5,3); $ans\=sig(3,sqrt(2*$d/(9.8*sin($angle*3.14/180)))+(1/($coeff_kinetic*9.8)*sqrt(2*$d*9.8*sin($angle*3.14/180))))@ qu.2.2.hint.1=To approach the problem, first draw a free body diagram and label all the forces acting on the block. Make sure to include a coordinate system. Then, apply Newton’s second law of motion in the x and y directions. This problem contains two parts, because first the block slides down a frictionless incline and then moves along a rough floor, in which case you do need to account for friction.@ qu.2.2.hint.2=When the block is on the incline (assuming that the x axis is directed along the incline):
ΣFx = mg (sin α) = ma
ΣFy = N – mg (cos α) = 0

When the block is moving along the floor (assuming that the x axis is directed along the floor):
ΣFx = - μkN = ma
ΣFy = N – mg = 0@ qu.2.2.hint.3=Use the two systems of equations to find the acceleration of the block on each of the intervals.@ qu.2.2.comment= 1. To approach the problem, first draw a free body diagram and label all the forces acting on the block. Make sure to include a coordinate system. Then, apply Newton’s second law of motion in the x and y directions. This problem contains two parts, because first the block slides down a frictionless incline and then moves along a rough floor, in which case you do need to account for friction.

2. When the block is on the incline (assuming that the x axis is directed along the incline):
ΣFx = mg (sin α) = ma
ΣFy = N – mg (cos α) = 0

When the block is moving along the floor (assuming that the x axis is directed along the floor):
ΣFx = - μkN = ma
ΣFy = N – mg = 0

3. Use the two systems of equations to find the acceleration of the block on each of the intervals.

4. Use the equation x = v0 t + (1/2)at 2 to find the time the block spent on the inclined surface.
Also, you must find the velocity of the block as it leaves the incline. This will be the initial velocity of the block as it reaches the rough horizontal floor and begins to decelerate. Find this value using the following equation: v = at.
To find the time the block spent decelerating on the rough floor, use the equation v = v0 + at, where v0 is the velocity of the block as it leaves the inclined surface and v = 0 m/s (since the block eventually comes to a stop).@ qu.2.3.mode=Randomized Formula@ qu.2.3.editing=useHTML@ qu.2.3.name=Block - Incline 1@ qu.2.3.question=A block of mass m = $m g (initially at rest at point A) starts to slide down a frictionless incline AB. When it reaches the rough surfaced segment BC, it begins to decelerate. Suppose that at point C the block is moving with a speed of $v m/s. What is the magnitude of the force that you must apply to the block in the direction of its motion at point C in order to keep it moving with constant speed? Assume that the coefficient of kinetic friction between the rough surface and the block is equal $coeff_kinetic.

@ qu.2.3.answer=$ans N (1 ? 0.05)@ qu.2.3.random= $m\=rand(300,900,3); $v\=rand(0.2,2,3); $coeff_kinetic\=rand(0.3,0.7,3); $ans\=sig(3,$coeff_kinetic*$m*9.8/1000)@ qu.2.3.hint.1=When an object moves with constant velocity, the acceleration of the object is zero, and the net force acting on the object is also zero. Thus, in this problem, the key idea is that the external force applied to the block at point C must compensate the decelerating force acting on it at that point.@ qu.2.3.comment= 1. When an object moves with constant velocity, the acceleration of the object is zero, and the net force acting on the object is also zero. Thus, in this problem, the key idea is that the external force applied to the block at point C must compensate the decelerating force acting on it at that point.

2. The only decelerating force that acts on the block at point C is the kinetic frictional force (due to the friction between the block and the floor).
If we direct the x axis horizontally and the y axis vertically, Newton's second law for the block (at point C) can be written down as follows:
ΣFx = μkN = ma, where a is the magnitude of deceleration of the block on the rough floor.
ΣFy = N – mg = 0.
Thus, the decelerating force acting on the block equals Fdecel = μkmg and the external force must be equal in magnitude to Fdecel but opposite in direction. @ qu.2.4.mode=Randomized Formula@ qu.2.4.editing=useHTML@ qu.2.4.name=Pulling a Box@ qu.2.4.question=Calculate the magnitude of the force F required to pull a box of mass m = $m kg with a constant speed of $v m/s if the coefficient of kinetic friction between the box and the surface is $coeff_kinetic. Assume that the pulley is massless and frictionless.

@ qu.2.4.answer=$ans N (1 ? 0.05)@ qu.2.4.random= $v\=rand(2,5,3); $m\=rand(3,10,3); $coeff_kinetic\=rand(0.2,0.7,3); $ans\=sig(3,$coeff_kinetic*$m*9.8)@ qu.2.4.hint.1=When an object moves with constant velocity, the acceleration of the object is zero, and the net force acting on the object is also zero. Thus, in this problem, the key idea is that the external force applied to the box must compensate the decelerating force acting on it.@ qu.2.4.comment= 1. When an object moves with constant velocity, the acceleration of the object is zero, and the net force acting on the object is also zero. Thus, in this problem, the key idea is that the external force applied to the box must compensate the decelerating force acting on it.

2. The only decelerating force that acts on the box is the kinetic frictional force (due to the friction between the box and the table).
If we direct the x axis horizontally and the y axis vertically, Newton's second law for the box can be written down as follows:
ΣFx = μkN - T = 0, where T is the tension force of the rope.
Because we assume that the rope does not stretch, the magnitude of T is equal to that of F.
ΣFy = N – mg = 0.
Thus, the magnitude of the force required to pull the box with constant speed equals: F = μkmg. @ qu.3.topic=Ch 6, Sec 3 - Static Frictional Force@ qu.3.1.mode=Randomized Formula@ qu.3.1.editing=useHTML@ qu.3.1.name=Block - Verge of Sliding@ qu.3.1.question=A block is placed on an incline as shown in the picture below. If you were able to change the angle of the incline from 0° to 90°, what would be the measure of the angle a, in degrees (deg), at which the block is on the verge of sliding in the direction of the x-axis? Suppose the coefficient of static friction between the block and the surface of the incline to be $coeff_kinetic.

@ qu.3.1.answer=$ans deg (1 ? 0.05)@ qu.3.1.random= $coeff_kinetic\=rand(0.2,0.7,3); $ans\=sig(3,180/3.14*arctan($coeff_kinetic))@ qu.3.1.hint.1=To approach the problem, first draw a free body diagram and label all the forces acting on the block.@ qu.3.1.hint.2=Applying Newton’s second law to the block, we obtain the following two equations (assuming that the x axis is directed as shown in the picture, and the y axis is perpendicular to the incline):
ΣFx = mg (sin α) – μs N = 0
ΣFy = N – mg (cos α) = 0@ qu.3.1.hint.3= For the block to be on the verge of sliding, the static frictional force must be at its maximum, the point at which any additional force will cause movement. The maximum value of static frictional force will be that for which the frictional force just barely balances the component of the force due to gravity along the direction of movement.@ qu.3.1.comment= 1. Applying Newton’s second law to the block, we obtain the following two equations (assuming that the x axis is directed as shown in the picture, and the y axis is perpendicular to the incline):
ΣFx = mg (sin α) – μs N = 0
ΣFy = N – mg (cos α) = 0

2. For the block to be on the verge of sliding, the static frictional force must be at its maximum, the point at which any additional force will cause movement. The maximum value of static frictional force will be that for which the frictional force just barely balances the component of the force due to gravity along the direction of movement.

3. Thus, mg (sin α) = μs mg (cos α), which simplifies to tan α = μs@ qu.3.2.mode=Randomized Formula@ qu.3.2.editing=useHTML@ qu.3.2.name=Two Connected Boxes@ qu.3.2.question=Two large identical boxes, initially at rest, are placed on a rough surface and are connected by a massless durable cord. The force F applied to the cord at its middle causes the cord to stretch slightly so that the two segments, AB and AC, form an angle q with the horizontal. Find the mass m of each of the boxes if the coefficient of static friction between the boxes and the rough surface is $coeff_static, the angle q is $angle°, and the force F required to start the boxes moving is $force N.

@ qu.3.2.answer=$ans kg (1 ? 0.05)@ qu.3.2.random= $coeff_static\=rand(0.2,0.9,3); $force\=rand(500,900,3); $angle\=rand(3,8,3); $ans\=sig(3,$force/(2*9.8)+$force*cos($angle*3.14/180)/(2*sin($angle*3.14/180)*9.8*$coeff_static))@ qu.4.topic=Ch 6, Sec 4 - Connected Objects@ qu.4.1.mode=Randomized Formula@ qu.4.1.editing=useHTML@ qu.4.1.name=Three Connected Blocks@ qu.4.1.question=Three blocks connected by a massless string are being pulled by a force F = $force N. Find the tension force $tension_force if the masses of the blocks m1, m2, and m3 are equal $m1 kg, $m2 kg, and $m3 kg respectively. Assume there is no friction.

@ qu.4.1.answer=$ans N (1 ? 0.05)@ qu.4.1.random= $force\=rand(20,40,3); $integer\=rint(4); $m1\=rand(4,8,3); $k_big\=rand(0.4,0.9,3); $k_small\=rand(0.2,0.4,3); $m2\=sig(3,$m1*(1+$k_big)); $m3\=sig(3,$m1*(1-$k_small)); $a\=$force/($m1+$m2+$m3); $tension_force\=switch($integer,"T1","T2","T3","T4"); $ans1\=switch($integer,$a*$m1,$a*$m1,$m2*$a+$m1*$a,$m2*$a+$m1*$a); $ans\=sig(3,$ans1)@ qu.4.2.mode=Randomized Formula@ qu.4.2.editing=useHTML@ qu.4.2.name=2 Blocks, 1 Pulley (Combo)@ qu.4.2.question=A block on a frictionless surface is connected by a massless string over a massless, frictionless pulley to another block that is hanging vertically. Knowing that the blocks are initially at rest and that m1 = $m1 kg and m2 = $m2 kg, $qu.

@ qu.4.2.answer=$ans $units (1 ? 0.05)@ qu.4.2.random= $integer\=rint(5); $k\=rand(0.2,0.7,3); $m1\=rand(3,10,3); $m2\=sig(3,$m1*(1-$k)); $t\=rand(1,4,3); $qu\=switch($integer,"find the magnitude of acceleration of the larger block","find the magnitude of acceleration of the smaller block","find the tension in the cord","find the distance that the smaller block travels in $t s","find the distance that the larger block travels in $t s"); $units\=switch($integer,"m/s^2","m/s^2","N","m","m"); $ans1\=switch($integer,$m2*9.8/($m1+$m2),$m2*9.8/($m1+$m2),$m1*$m2*9.8/($m1+$m2),0.5*$m2*9.8/($m1+$m2)*($t)^2,0.5*$m2*9.8/($m1+$m2)*($t)^2); $ans\=sig(3,$ans1)@ qu.4.3.mode=Randomized Formula@ qu.4.3.editing=useHTML@ qu.4.3.name=2 Blocks, 2 Inclines@ qu.4.3.question=Two blocks located on inclined frictionless surfaces are connected by a massless cord over a massless and frictionless pulley (see picture). Find the magnitude of acceleration of the $smaller block if m1 = $m1 kg, m2 = $m2 kg, a = $alpha°, and b = $beta°.

@ qu.4.3.answer=$ans m/s^2 (1 ? 0.05)@ qu.4.3.random= $smaller\=switch(rint(2),"smaller","larger"); $k\=rand(0.3,0.7); $k_angle\=rand(0.4,0.7); $m1\=rand(1,4,3); $m2\=sig(3,$m1*(1+$k)); $alpha\=rand(30,65,3); $beta\=sig(3,$alpha(1-$k_angle)); $ans\=sig(3,abs(($m1*sin($alpha*3.142/180)-$m2*sin($beta*3.142/180))/($m1+$m2)*9.8))@ qu.4.4.mode=Randomized Formula@ qu.4.4.editing=useHTML@ qu.4.4.name=Pulleys -- System@ qu.4.4.question=Given m1 = $m1 kg and m2 = $m2 kg, find the magnitude of acceleration of the mass $mass in the system shown below. Assume that both the rope and the pulleys are massless and assume there is no friction.

@ qu.4.4.hint.1=To solve this problem, you need to apply the Newton's second law of motion to different bodies of the system shown in the picture.@ qu.4.4.hint.2=Write down a system of four equations that contains the following expressions: @ qu.4.4.hint.3=In order to apply Newton's second law of motion, we need to label all the forces acting on the considered bodies. We also need to pick a direction of the axis that we will project the vectors on (see picture). Because the directions of accelerations of the blocks are not given, we will assume that they are positive. The final answer will determine the actual directions.
The equation for Newton's second law for the block m1 is written as follows:
,
where a1 is the acceleration of mass m1 and T1 is the tension in the rope. Since we assume that the rope is massless and that it does not stretch, the tension in it is the same along the length.
Similarly, the equation for Newton's second law for the block m2 looks as follows:
,
where a2 is the acceleration of mass m2 and T2 is the tension in the rope connecting m2 with the respective pulley.@ qu.4.4.hint.4=Now we will write down the expression for Newton's second law for the pulley directly connected to m2.

Because we assumed the pulley to be massless, we get T2 = 2T1.@ qu.4.4.hint.5=Finally, we need to establish a relationship between the accelerations of the blocks.
Let us consider instantaneous displacement of the two blocks. Suppose that m1 moves a distance Dx1 over a period of time Dt, then m2 would move half that distance in the opposite direction. In other words, Dx1 = - 2Dx2. Since both displacements took place over the same infinitesimally small interval of time Dt, we may state the following:
,
where v1 and v2 are instantaneous velocities of m1 and m2 respectively.@ qu.4.4.hint.6=So we obtained the following system of four equations with four unknowns:

Find the answer by solving this system for a desired quantity.@ qu.4.4.answer=$ans m/s^2 (1 ? 0.05)@ qu.4.4.random= $integer\=rint(2); $mass\=switch($integer,"m1","m2"); $k\=rand(0.3,0.7); $m1\=rand(1,4,3); $m2\=sig(3,$m1*(2+$k)); $ans1\=switch($integer,2*9.8*((2*$m1-$m2)/(4*$m1+$m2)),9.8*((2*$m1-$m2)/(4*$m1+$m2))); $ans2\=sig(3,$ans1); $ans\=abs($ans2)@ qu.4.5.mode=Randomized Formula@ qu.4.5.editing=useHTML@ qu.4.5.name=Pulleys -- System 2@ qu.4.5.question=Find the tension in the rope if m1 = $m1 kg and m2 = $m2 kg. Assume that both the rope and the pulleys are massless and assume there is no friction.

@ qu.4.5.answer=$ans N (1 ? 0.05)@ qu.4.5.random= $k\=rand(0.3,0.7); $m1\=rand(1,4,3); $m2\=sig(3,$m1*(2+$k)); $ans\=sig(3,3*9.8*$m1*$m2/(4*$m1+$m2))@ qu.5.topic=Ch 6, Sec 5 - Springs and Strings@ qu.5.1.mode=Randomized Formula@ qu.5.1.editing=useHTML@ qu.5.1.name=Spring -- Static@ qu.5.1.question=A spring with a force constant k = $k N/m is attached to a block of mass m = $m kg that is initially at rest. The force F applied to the spring stretches it by Dx m. Given a = $a°, find Dx corresponding to the minimum force F that has a large enough magnitude to start the block moving. The coefficient of static friction between the block and the floor is $k_static.

@ qu.5.1.answer=$ans m (1 ? 0.05)@ qu.5.1.random= $k\=rand(75,200,3); $m\=rand(5,15,3); $k_static\=rand(0.1,0.8,3); $a\=rand(10,65,3); $ans\=sig(3,($k_static*$m*9.8)/($k*cos($a*3.14/180)+$k*$k_static*sin($a*3.14/180)))@ qu.5.2.mode=Randomized Formula@ qu.5.2.editing=useHTML@ qu.5.2.name=Spring -- Static 2@ qu.5.2.question=A spring firmly attached to a block of mass m = $m kg is used to push the block along a flat surface. What must the force constant of the spring be if the spring compresses by Dx = $dx cm at the moment the block begins to move? The coefficient of static friction between the block and the surface is $k_static.

@ qu.5.2.answer=$ans N/m (1 ? 0.05)@ qu.5.2.random= $k_static\=rand(0.1,0.8,3); $dx\=rand(1,6,3); $m\=rand(2,5,3); $ans\=sig(3,$k_static*$m*9.8/($dx/100))@ qu.5.3.mode=Randomized Formula@ qu.5.3.editing=useHTML@ qu.5.3.name=2 Strings (1 Horizontal)@ qu.5.3.question=A block of mass m = $m kg is supported by two strings as shown in the picture below. Find the tension in the rope $qu if the angle a is equal to $a°.

@ qu.5.3.answer=$ans N (1 ? 0.05)@ qu.5.3.random= $integer\=rint(2); $m\=rand(2,7,3); $a\=rand(20,50,3); $qu\=switch($integer,"A","B"); $ans1\=switch($integer,9.8*$m/sin($a*3.14/180),9.8*$m/tan($a*3.14/180)); $ans\=sig(3,$ans1)@ qu.5.4.mode=Randomized Formula@ qu.5.4.editing=useHTML@ qu.5.4.name=2 Supporting Ropes@ qu.5.4.question=A $box is supported by two strings as shown in the picture below. What is the mass of the box if the sum of the magnitudes of the tension forces of the strings A and B is equal to $f_sum N and the angles a and b are equal $a° and $b° respectively?

@ qu.5.4.answer=$ans kg (1 ? 0.05)@ qu.5.4.random= $box\=switch(rint(3),"box","block","body"); $k\=rand(0.3,0.7); $b\=rand(30,65,3); $a\=sig(3,$b(1-$k)); $f_sum\=rand(20,99,3); $Ta\=$f_sum-($f_sum*cos($a*3.14/180))/(cos($b*3.14/180)+cos($a*3.14/180)); $Tb\=$f_sum*cos($a*3.14/180)/(cos($a*3.14/180)+cos($b*3.14/180)); $ans1\=1/9.8*($Ta*sin($a*3.14/180)+$Tb*sin($b*3.14/180)); $ans\=sig(3,$ans1)@ qu.5.5.mode=Randomized Formula@ qu.5.5.editing=useHTML@ qu.5.5.name=2 Identical Blocks@ qu.5.5.question=Two identical blocks of unknown mass m are supported by three strings as shown in the picture. Find the tension in the string C if the tension of string A and the angles a and g are equal $Ta N, $alpha° and $gamma° respectively.

@ qu.5.5.answer=$ans N (1 ? 0.05)@ qu.5.5.random= $Ta\=rand(50,99,3); $alpha\=rand(45,60,3); $gamma\=rand(50,75,3); $ans\=sig(3,$Ta*cos($alpha*3.14/180)/cos($gamma*3.14/180))@ qu.6.topic=Ch 6, Sec 6 - Circular Motion@ qu.6.1.mode=Randomized Formula@ qu.6.1.editing=useHTML@ qu.6.1.name=Cornering@ qu.6.1.question=What is the maximum speed that $an_automobile can have going around a corner of radius R = $r m without skidding. The coefficient of static friction between the tires and the road is $k_static.@ qu.6.1.answer=$ans m/s (1 ? 0.05)@ qu.6.1.random= $an_automobile\=switch(rint(3),"an automobile","a truck","a car"); $r\=rand(100,350,3); $k_static\=rand(0.1,0.8,3); $ans\=sig(3,(9.8*$r*$k_static)^(1/2))@ qu.6.2.mode=Randomized Formula@ qu.6.2.editing=useHTML@ qu.6.2.name=Cornering [Banked]@ qu.6.2.question=A $car is traveling on a banked circular road of radius $r m. Find the magnitude of its linear velocity that will allow the $car to round the corner without any assistance from friction if the banking angle is $angle°.@ qu.6.2.answer=$ans m/s (1 ? 0.05)@ qu.6.2.random= $car\=switch(rint(4),"racecar","car","vehicle","truck"); $angle\=rand(10,30,3); $r\=rand(70,300,3); $ans\=sig(3,(9.8*$r*tan($angle*3.14/180))^(1/2))@ qu.6.3.mode=Randomized Formula@ qu.6.3.editing=useHTML@ qu.6.3.name=Loop the Loop 1@ qu.6.3.question=Consider the motion of the "loop the loop" roller coaster shown below. What is the magnitude of the force exerted on the tracks by the roller coaster $qu if the magnitude of its linear velocity at that point is $v m/s, its mass is $m kg, and the radius of the circular tracks is $r m?

@ qu.6.3.answer=$ans N (1 ? 0.05)@ qu.6.3.random= $m\=rand(4000,6500,3); $v\=rand(25,45,3); $r\=rand(30,40,3); $integer\=rint(2); $qu\=switch($integer,"at the top of the loop (at point A)","at the bottom of the loop (at point B)"); $ans1\=switch($integer,$m*($v)^2/$r-9.8*$m,$m*($v)^2/$r+9.8*$m); $ans\=sig(3,$ans1)@ qu.6.4.mode=Randomized Formula@ qu.6.4.editing=useHTML@ qu.6.4.name=Loop the Loop 2 {Zero Normal}@ qu.6.4.question=Consider the motion of the "loop the loop" roller coaster shown below. What must be the magnitude of linear velocity of the roller coaster at the top of the loop (at point A) in order for the force exerted by the roller coaster on the tracks to be zero? Assume that the radius of the circular tracks is $r m.

@ qu.6.4.hint.1=Apply Newton's second law of motion to the roller coaster to solve this problem.@ qu.6.4.hint.2= At the top of the loop, the roller coaster is acted on by two forces: the normal force from the tracks (N) and the gravitational force (mg) (see close-up).@ qu.6.4.hint.3= Let us choose the axis of projection to be directed radially toward the center of the loop. Then, Newton's second law projected on that axis is written as follows:

Since centripetal acceleration can be expressed as
,
we get:
@ qu.6.4.hint.4= According to Newton's third law, the magnitude of the normal force N is equal to that of the force exerted on the tracks by the roller coaster. Thus, if N = 0 N, the force exerted by the roller coaster on the tracks is equal to zero.@ qu.6.4.hint.5=Likewise,
.
Substitute respective numerical values and calculate the magnitude of linear velocity of the roller coaster in m/s.@ qu.6.4.answer=$ans m/s (1 ? 0.05)@ qu.6.4.random= $r\=rand(25,50,3); $ans\=sig(3,(9.8*$r)^(1/2))@ qu.6.5.mode=Randomized Formula@ qu.6.5.editing=useHTML@ qu.6.5.name=Pendulum@ qu.6.5.question=A ball is attached to a massless string of length l = $r m to form a pendulum. At some point in time the pendulum is put in motion. During the oscillation, the ball follows a circular path with the center at point O. When the ball is at point P, the tension of the string is $T N and the magnitude of the ball's linear velocity at that point is $v m/s. Find the mass of the ball.

@ qu.6.5.answer=$ans kg (1 ? 0.05)@ qu.6.5.random= $r\=rand(1,2.5,3); $T\=rand(60,200,3); $v\=rand(10,25,3); $ans\=sig(3,$T/(9.8+($v)^2/$r))@