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HINTS FOR PROBLEM SET #7
Problem Set #7 – Problem 1 Part a) You are given information about the velocity a truck has at different points during its displacement as it slows down. You are not asked to find the acceleration. Therefore you should use the Work Energy Theorem. T1 + W1→2 = T2 where T1 is the Kinetic Energy of the system at position 1. W1→2 = the work done by all the forces acting on the system during the displacement between position 1 and position 2. T2 is the Kinetic Energy of the system at position 2.
Remember that the subscripts in this formula refer to positions of the particle.
Let position 1 be the position when the truck has the velocity of 65 mph. Let position 2 be the position when it has 40 mph.
Calculating the Kinetic Energy at each of these positions is relatively straightforward: T1 = 1/2 m v12 T2 = 1/2 m v22 Be careful to convert the velocity from mph to ft/s. Also convert the weight to a mass. (Just divide it by 32.2 ft/s2.)
The get the work done by all the forces acting during the displacement, you need to calculate the work done by each force and then add them up.
It is essential to account for all of the forces acting. To be sure that you account for all of them draw a FBD of the truck between the two positions. The forces are: - the weight, W - the normal force, N - the frictional force, f
Work done by the weight = W x (vertical displacement)
Work done by the normal force is zero.
Work done by the frictional force = f x (displacement) Note: This assumes that friction is constant over the displacement which may or may not be true. However, since you only need the average frictional force this formula will give it to you. Furthermore, this frictional force, f, is the average braking force that you are asked to find.
Now add up all of these work terms and plug into the W-E Theorem. Solve for f, the average frictional force.
Part b) You need to find the average coupling force between the cab and the trailer that acts during the displacement of the truck. Follow essentially the same procedure as in part a.
Apply the W-E Theorem between position 1 and position 2, however, this time draw a FBD of the trailer only. An addition force on this FBD will be the force in the coupling, Fc.
The calculations of T1 and T2 are the same as above except the mass is for the trailer only.
The calculation of the work of each force acting during the displacement is almost the same as part (a).
Work done by the weight = W x (vertical displacement) where in this case W is for the trailer only.
Work done by the normal force is zero. (Since the normal force acts perpendicular to the displacement.)
Work done by the frictional force = f x (displacement) Again this friction force, f, represents the average frictional force, or the average braking force. In this case you are told that the trailer supplies 70% of the braking force. So this force, f, is 70% of the force you found in part (a).
Work done by the coupling force = Fc x (displacement)
Now add up all of these work terms and plug into the W-E Theorem. Solve for Fc, the average coupling force. |
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Problem Set #7 –
Problem 2 First, you need to recognize that the motions of blocks A and B are dependent. The system has two motions and one degree of freedom. The velocity of block A is equal to that of block B but in the opposite direction. vA = -vB = v Since you are told something about displacements, and are not worried about finding acceleration, use the Work Energy Theorem. T1 + W1→2 = T2 where T1 is the Kinetic Energy of the system at position 1. W1→2 = the work done by all the forces acting on the system during the displacement between position 1 and position 2. T2 is the Kinetic Energy of the system at position 2.
In this problem you need to split the displacement into two separate intervals and apply the W-E Theorem to each interval.
The first displacement interval is from the initial
position of the system to the position of the system when collar C is removed from the top of block A (call this position 2.) The W-E Theorem will look like this: T1 +
W1→2 = T2 is from the position of the system when collar C is removed from the top of block A (this is position 2) to the position of the system just before A hits the floor (call this position 3.) The W-E Theorem will look like this: T2
+ W2→3 = T3 For each interval you will - first apply the W-E Theorem to block A - then apply the W-E Theorem to block B - then add these two equations to
get one equation
First displacement interval: Draw the FBD for block A (with collar C on it). Draw the FBD for block B. In each of the FBDs there will be a tension force, T, from the rope. The tension forces will be the same in each FBD because the pulleys are frictionless.
Apply the W-E Theorem for each FBD between position 1 and position 2. Add the two equations and you will get one equation where the terms with T disappear. (The terms with T cancel out because the tension does negative work on block A and positive work on block B. They are traveling through the same displacement and so the calculation of the work will be same magnitude but opposite sign.)
Use this equation to solve for the velocity at the end of the first interval.
Second displacement interval: The procedure is essentially the same. Draw two FBDs: One for block A (alone this time) and another for block B. Again, there will be a tension force, T, from the rope in each diagram. The tension forces in both of these diagrams have the same magnitude. Apply the W-E Theorem for each FBD between position 2 and position 3. Add the two equations to get one equation where the terms with T disappear.
Use this equation to solve for the velocity at the end of the second interval.
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Problem Set #7 –
Problem 3 You are given a system of a block and spring. You know the position and the velocity of the block at the beginning of the motion. (The velocity is zero since it is "released from rest". You can easily calculate the position from Hook's Law.) You are asked to find the velocity at another position and also the maximum velocity achieved.
Using the Work-Energy Theorem is the obvious way to go since you want to find velocities and you know something about the positions where you want to find these velocities. (And you are not worried about finding accelerations.) To use the W-E Theorem you need to clearly identify the positions:
Part a) Apply the W-E Theorem between position 1 and position 2.
Part b) Apply the W-E Theorem between position 2 and position 3. But here is the hard part of the problem. Position 3 is a bit tricky. Where exactly is it? To find out exactly where position 3 is, use the fact that the maximum velocity corresponds to the point when the acceleration is zero. Why is this? Because from calculus you know that the maximum (or minimum) of a function occurs when the slope is zero. Since a = dv/dt, the slope of the velocity curve is equal to the acceleration. So the velocity will be a maximum (or minimum) when the acceleration is zero. Zero acceleration corresponds to when the net force is zero. The only forces acting in the x direction are the spring force and friction. hence position 3 is where these two forces cancel each other out. Hence, position 3 by finding out when the friction force = the elastic spring force. Solve this equation for x:
This value of x is position 3.
Now that you know exactly where position 3 is, you can plug into the W-E Theorem and solve for v3. This is the maximum velocity. |