Physics 2211 Quiz #2 Solutions Spring 2008 Unless otherwise ...
Physics 2211 Spring 2008
Quiz #2
Solutions
Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) A block of mass m accelerates up a frictionless incline that makes an angle θ with the horizontal due to a horizontal applied force of magnitude A. The coefficient of static friction between the block and the ramp is µs , while the coefficient of kinetic friction between the block and the ramp is µk . What is the acceleration magnitude of the block, in terms of any or all of m, θ, A, µs , µk , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
Announced during quiz: the incline is not frictionless. Use Newton’s Second Law. Sketch a Free Body Diagram. Choose a coordinate system. In this case, a coordinate system has been chosen so that the x axis points in the direction of the acceleration. Resolve any forces that do not lie along an axis into components. Write out Newton’s Second Law for each direction. First, the y direction, from which the normal force, n, will be determined: X
Fy = n − Ay − wy = may = 0
⇒
n = A sin θ + mg cos θ
where A is the magnitude of the applied force and w is the magnitude of the weight force. Next, the x direction: X
Fx = Ax − wx − fk = max
⇒
max = A cos θ − mg sin θ − µk n
where fk is the force of kinetic friction. Substituting the expression for the normal force found from the y direction: max = A cos θ − mg sin θ − µk (A sin θ + mg cos θ) So
ax =
A cos θ − mg sin θ − µk A sin θ − µk mg cos θ m
or
ax =
A (cos θ − µk sin θ) − mg (sin θ + µk cos θ) m
Quiz #2 Solutions Page 1 of 6
II. (16 points) A ping-pong ball of mass m is thrown straight downward with a speed that is twice its terminal speed. What is the magnitude of its acceleration upon release, in terms of any or all of m and physical or mathematical constants? (On Earth, do NOT neglect drag!) . . . . . . . . . . . . . . . . . . . . . . . Use Newton’s Second Law. Sketch a Free Body Diagram. Choose a coordinate system. In this case, a coordinate system has been chosen so that the x axis points in the direction of the acceleration when the speed is twice the terminal speed. First, we’d like to know how the drag force is related to the weight force. Write out Newton’s Second Law for the situation where the speed, v, is the same as the terminal speed, vt . X
Fx = D − w = max = 0
⇒
D=w
So 2 1 4 Avt
= mg
or
2 1 2 CρAvt
= mg
Note that we’ve used D = 14 Av 2 , rather than D ≈ 14 Av 2 . The approximation only means that the factor may be slightly different from 41 . In this problem, all that matters is that this factor be a constant, such as 1 2 Cρ. The important thing to note here is that the drag force is proportional to the square of the terminal speed, and that it equals the weight force at the terminal speed. So, when the speed is twice the terminal speed X
Fx = D − w = max
⇒
2
max = 14 A (2vt ) − mg
⇒
max = 4
¡1
2 4 Avt
¢
− mg
From the situation where the speed was equal to the terminal speed, we found that 14 Avt2 = mg, so max = 4 (mg) − mg = 3mg
⇒
ax = 3g
1. (6 points) If positive is chosen upward in the problem above, which graph best represents the velocity of the ping-pong ball as a function of time? . . . . . . . . . . . . . . . . . . . . . . . Since the drag force upward is initially greater than the weight, the graph of velocity as a function of time must have a positive initial slope. This slope must approach zero as the speed approaches the terminal speed (which is not zero).
Quiz #2 Solutions Page 2 of 6
III. (16 points) A wooden block of mass M is launched up a ramp that is inclined at an angle θ with the horizontal. The initial speed of the block is v0 . The coefficient of static friction between the block and the ramp is µs , while the coefficient of kinetic friction between the block and the ramp is µk . Neither µs nor µk is zero. How far along the ramp does the block slide before it stops, in terms of any or all of M , θ, v0 , µs , µk , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
This is a two-part problem. First, Newton’s Second Law will be used to find the constant acceleration of the block. Then the constant acceleration kinematics relationships can be used to determine the distance the block slides. Sketch a Free Body Diagram to start with Newton’s Second Law. Choose a coordinate system. In this case, a coordinate system has been chosen so that the x axis points in the direction of the acceleration. Resolve any forces that do not lie along an axis into components. Write out Newton’s Second Law for each direction. Forces in the y direction can be used to find the normal force, n. X
Fy = n − wy = M ay = 0
⇒
n = M g cos θ
where w is the weight of the block. In the x direction X
Fx = fk + wx = M ax
⇒
M ax = µk n + M g sin θ
where fk is the force of kinetic friction. Substituting the expression for the normal force found above M ax = µk (M g cos θ) + M g sin θ
⇒
ax = g (µk cos θ + sin θ)
Next, this acceleration can be used in the kinematics step of the problem. 2
x = x0 + v0x ∆t + 12 ax (∆t)
and
vx = v0x + ax ∆t
Solving the velocity equation for ∆t and substituting into the position equation will eliminate time from the two equations: vx − v0x ∆t = ax
µ ⇒
x − x0 = v0x
vx − v0x ax
¶
µ +
1 2 ax
vx − v0x ax
¶2 ⇒
2 2ax (x − x0 ) = vx2 − v0x
Solve for the displacement. Note that vx is zero, as the block stops at its greatest distance up the ramp. The expression for the acceleration found can be substituted for ax . x − x0 =
2 2 −v0x −v0x = 2ax 2g (µk cos θ + sin θ)
Note that the choice of coordinate systems makes this displacement negative, while the distance is best expressed as its magnitude. 2 v0x |x − x0 | = 2g (µk cos θ + sin θ)
Quiz #2 Solutions Page 3 of 6
2. (6 points) In the problem above, assume the block does not remain at the point where it stops sliding up the ramp, but slides back down. How does the speed of the block when it reaches its launch point on the way down, vdown , compare to the launch speed v0 ? . . . . . . . . . . . . . . . . . . . . . . . If there were no friction, then vdown would be equal to v0 . With friction, however, the block does not slide as far up the ramp, and so is sliding back down from a lesser height. Additionally, the friction force reduces the magnitude of the acceleration as the block slides back down, so vdown < v0
Quiz #2 Solutions Page 4 of 6
3. (10 points) Sue is pulling on the crate with a force magnitude FP . The crate has mass m, and coefficient of static friction µs with the level ground. Because of this static friction, the crate does not move. Sue gets tired, and reduces her force to FP /2. What is the magnitude of the static friction force fs on the crate, now that Sue has reduced her force? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . The block is in static equilibrium, so the net force on it is zero. The horizontal forces of static friction and Sue’s pull must be opposite in direction and equal in magnitude. fs = FP /2
4. (10 points) Bob is standing on an ordinary bathroom scale in an elevator. He glances down at the scale, and notes that the reading is greater than his weight. Consider the following situations: (i) The elevator is travelling upward. (ii) The elevator is travelling downward. (iii) The elevator is stationary. Which situation or situations could be true at the instant Bob glances at the scale? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . Since the scale reading (normal force) is greater than Bob’s weight, the net force on him must be upwards, and his acceleration must be upwards. He could be travelling upward with increasing speed, downward with decreasing speed, or he could be momentarily stationary as the elevator changes direction from downward to upward. The situation when Bob glances at the scale, then, could be Any of situations i, ii, or iii.
Quiz #2 Solutions Page 5 of 6
5. (10 points) A block of mass m is placed on a horizontal planar sheet that is pivoted at one end. The coefficient of static friction between the block and the plane is µs , while that of kinetic friction is µk . The free end of the sheet is then raised very slowly. When the end is first raised, static friction between the block and the sheet keeps the block from moving. At a certain angle θ, however, the block begins to slide down the inclined sheet. How does the velocity of the block depend on time once it begins to move? (On Earth, let the positive direction be down the incline.) . . . . . . . . . . . . . . . . . . . . . . . The coefficient of kinetic friction between the block and the sheet must be less than the coefficient of static friction between them. At a particular angle θ, then, the force of kinetic friction on the block as it slides down the plane must be less than the force of static friction on it before it starts to slide. There will be a constant net force down the plane (that is, in the positive direction) and therefore a constant positive acceleration. The graph of velocity as a function of time must have a constant positive slope. OR The coefficient of kinetic friction between the block and the sheet must be less than the coefficient of static friction between them. The force of kinetic friction on the block as it slides down the plane must be less than the force of static friction on it before it starts to slide. As the sheet continues to be raised, there will be an increasing net force down the plane (that is, in the positive direction) and therefore an increasing positive acceleration. The graph of velocity as a function of time must have an increasing positive slope. Full credit was issued for either interpretation.
6. (10 points) The truck can accelerate to the right over level ground without causing the crate on the bed to slide. If the acceleration is too great, however, the crate will slide and fall off the back of the truck. If the crate does slide and fall off the back, what horizontal force, if any, is acting on the crate while it slides, and in what direction? . . . . . . . . . . . . . . . . . . . . . . . Since it is possible for the truck to accelerate without causing the crate to slide, there must be a frictional force between the truck and the crate. When the crate does slide, that frictional force must be kinetic. When the crate slides toward the rear of the truck, kinetic friction opposes this and must be toward the right. Or, when the truck accelerates toward the right, the crate also accelerates toward the right even if it slides, so once again A force of kinetic friction acts toward the right.
Quiz #2 Solutions Page 6 of 6
Quiz #2
Solutions
Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) A block of mass m accelerates up a frictionless incline that makes an angle θ with the horizontal due to a horizontal applied force of magnitude A. The coefficient of static friction between the block and the ramp is µs , while the coefficient of kinetic friction between the block and the ramp is µk . What is the acceleration magnitude of the block, in terms of any or all of m, θ, A, µs , µk , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
Announced during quiz: the incline is not frictionless. Use Newton’s Second Law. Sketch a Free Body Diagram. Choose a coordinate system. In this case, a coordinate system has been chosen so that the x axis points in the direction of the acceleration. Resolve any forces that do not lie along an axis into components. Write out Newton’s Second Law for each direction. First, the y direction, from which the normal force, n, will be determined: X
Fy = n − Ay − wy = may = 0
⇒
n = A sin θ + mg cos θ
where A is the magnitude of the applied force and w is the magnitude of the weight force. Next, the x direction: X
Fx = Ax − wx − fk = max
⇒
max = A cos θ − mg sin θ − µk n
where fk is the force of kinetic friction. Substituting the expression for the normal force found from the y direction: max = A cos θ − mg sin θ − µk (A sin θ + mg cos θ) So
ax =
A cos θ − mg sin θ − µk A sin θ − µk mg cos θ m
or
ax =
A (cos θ − µk sin θ) − mg (sin θ + µk cos θ) m
Quiz #2 Solutions Page 1 of 6
II. (16 points) A ping-pong ball of mass m is thrown straight downward with a speed that is twice its terminal speed. What is the magnitude of its acceleration upon release, in terms of any or all of m and physical or mathematical constants? (On Earth, do NOT neglect drag!) . . . . . . . . . . . . . . . . . . . . . . . Use Newton’s Second Law. Sketch a Free Body Diagram. Choose a coordinate system. In this case, a coordinate system has been chosen so that the x axis points in the direction of the acceleration when the speed is twice the terminal speed. First, we’d like to know how the drag force is related to the weight force. Write out Newton’s Second Law for the situation where the speed, v, is the same as the terminal speed, vt . X
Fx = D − w = max = 0
⇒
D=w
So 2 1 4 Avt
= mg
or
2 1 2 CρAvt
= mg
Note that we’ve used D = 14 Av 2 , rather than D ≈ 14 Av 2 . The approximation only means that the factor may be slightly different from 41 . In this problem, all that matters is that this factor be a constant, such as 1 2 Cρ. The important thing to note here is that the drag force is proportional to the square of the terminal speed, and that it equals the weight force at the terminal speed. So, when the speed is twice the terminal speed X
Fx = D − w = max
⇒
2
max = 14 A (2vt ) − mg
⇒
max = 4
¡1
2 4 Avt
¢
− mg
From the situation where the speed was equal to the terminal speed, we found that 14 Avt2 = mg, so max = 4 (mg) − mg = 3mg
⇒
ax = 3g
1. (6 points) If positive is chosen upward in the problem above, which graph best represents the velocity of the ping-pong ball as a function of time? . . . . . . . . . . . . . . . . . . . . . . . Since the drag force upward is initially greater than the weight, the graph of velocity as a function of time must have a positive initial slope. This slope must approach zero as the speed approaches the terminal speed (which is not zero).
Quiz #2 Solutions Page 2 of 6
III. (16 points) A wooden block of mass M is launched up a ramp that is inclined at an angle θ with the horizontal. The initial speed of the block is v0 . The coefficient of static friction between the block and the ramp is µs , while the coefficient of kinetic friction between the block and the ramp is µk . Neither µs nor µk is zero. How far along the ramp does the block slide before it stops, in terms of any or all of M , θ, v0 , µs , µk , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
This is a two-part problem. First, Newton’s Second Law will be used to find the constant acceleration of the block. Then the constant acceleration kinematics relationships can be used to determine the distance the block slides. Sketch a Free Body Diagram to start with Newton’s Second Law. Choose a coordinate system. In this case, a coordinate system has been chosen so that the x axis points in the direction of the acceleration. Resolve any forces that do not lie along an axis into components. Write out Newton’s Second Law for each direction. Forces in the y direction can be used to find the normal force, n. X
Fy = n − wy = M ay = 0
⇒
n = M g cos θ
where w is the weight of the block. In the x direction X
Fx = fk + wx = M ax
⇒
M ax = µk n + M g sin θ
where fk is the force of kinetic friction. Substituting the expression for the normal force found above M ax = µk (M g cos θ) + M g sin θ
⇒
ax = g (µk cos θ + sin θ)
Next, this acceleration can be used in the kinematics step of the problem. 2
x = x0 + v0x ∆t + 12 ax (∆t)
and
vx = v0x + ax ∆t
Solving the velocity equation for ∆t and substituting into the position equation will eliminate time from the two equations: vx − v0x ∆t = ax
µ ⇒
x − x0 = v0x
vx − v0x ax
¶
µ +
1 2 ax
vx − v0x ax
¶2 ⇒
2 2ax (x − x0 ) = vx2 − v0x
Solve for the displacement. Note that vx is zero, as the block stops at its greatest distance up the ramp. The expression for the acceleration found can be substituted for ax . x − x0 =
2 2 −v0x −v0x = 2ax 2g (µk cos θ + sin θ)
Note that the choice of coordinate systems makes this displacement negative, while the distance is best expressed as its magnitude. 2 v0x |x − x0 | = 2g (µk cos θ + sin θ)
Quiz #2 Solutions Page 3 of 6
2. (6 points) In the problem above, assume the block does not remain at the point where it stops sliding up the ramp, but slides back down. How does the speed of the block when it reaches its launch point on the way down, vdown , compare to the launch speed v0 ? . . . . . . . . . . . . . . . . . . . . . . . If there were no friction, then vdown would be equal to v0 . With friction, however, the block does not slide as far up the ramp, and so is sliding back down from a lesser height. Additionally, the friction force reduces the magnitude of the acceleration as the block slides back down, so vdown < v0
Quiz #2 Solutions Page 4 of 6
3. (10 points) Sue is pulling on the crate with a force magnitude FP . The crate has mass m, and coefficient of static friction µs with the level ground. Because of this static friction, the crate does not move. Sue gets tired, and reduces her force to FP /2. What is the magnitude of the static friction force fs on the crate, now that Sue has reduced her force? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . The block is in static equilibrium, so the net force on it is zero. The horizontal forces of static friction and Sue’s pull must be opposite in direction and equal in magnitude. fs = FP /2
4. (10 points) Bob is standing on an ordinary bathroom scale in an elevator. He glances down at the scale, and notes that the reading is greater than his weight. Consider the following situations: (i) The elevator is travelling upward. (ii) The elevator is travelling downward. (iii) The elevator is stationary. Which situation or situations could be true at the instant Bob glances at the scale? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . Since the scale reading (normal force) is greater than Bob’s weight, the net force on him must be upwards, and his acceleration must be upwards. He could be travelling upward with increasing speed, downward with decreasing speed, or he could be momentarily stationary as the elevator changes direction from downward to upward. The situation when Bob glances at the scale, then, could be Any of situations i, ii, or iii.
Quiz #2 Solutions Page 5 of 6
5. (10 points) A block of mass m is placed on a horizontal planar sheet that is pivoted at one end. The coefficient of static friction between the block and the plane is µs , while that of kinetic friction is µk . The free end of the sheet is then raised very slowly. When the end is first raised, static friction between the block and the sheet keeps the block from moving. At a certain angle θ, however, the block begins to slide down the inclined sheet. How does the velocity of the block depend on time once it begins to move? (On Earth, let the positive direction be down the incline.) . . . . . . . . . . . . . . . . . . . . . . . The coefficient of kinetic friction between the block and the sheet must be less than the coefficient of static friction between them. At a particular angle θ, then, the force of kinetic friction on the block as it slides down the plane must be less than the force of static friction on it before it starts to slide. There will be a constant net force down the plane (that is, in the positive direction) and therefore a constant positive acceleration. The graph of velocity as a function of time must have a constant positive slope. OR The coefficient of kinetic friction between the block and the sheet must be less than the coefficient of static friction between them. The force of kinetic friction on the block as it slides down the plane must be less than the force of static friction on it before it starts to slide. As the sheet continues to be raised, there will be an increasing net force down the plane (that is, in the positive direction) and therefore an increasing positive acceleration. The graph of velocity as a function of time must have an increasing positive slope. Full credit was issued for either interpretation.
6. (10 points) The truck can accelerate to the right over level ground without causing the crate on the bed to slide. If the acceleration is too great, however, the crate will slide and fall off the back of the truck. If the crate does slide and fall off the back, what horizontal force, if any, is acting on the crate while it slides, and in what direction? . . . . . . . . . . . . . . . . . . . . . . . Since it is possible for the truck to accelerate without causing the crate to slide, there must be a frictional force between the truck and the crate. When the crate does slide, that frictional force must be kinetic. When the crate slides toward the rear of the truck, kinetic friction opposes this and must be toward the right. Or, when the truck accelerates toward the right, the crate also accelerates toward the right even if it slides, so once again A force of kinetic friction acts toward the right.
Quiz #2 Solutions Page 6 of 6
