Instructions: (62 points) Answer the following questions. SHOW ALL OF YOUR WORK. A B = A x B x + A y B y + A z B z = ( 1) + ( 1) ( 4) = 5

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1 AP Physics C Fall, 2016 Work-Energy Mock Exam Name: Answer Key Mr. Leonard Instructions: (62 points) Answer the following questions. SHOW ALL OF YOUR WORK. (12 pts ) 1. Consider the vectors A = 2 î + 3 ĵ 1 ˆk and B = 2 î + 1 ĵ 4 ˆk (a) (6 pts) Calculate the Dot Product A B Solution: 12 pts A B = A x B x + A y B y + A z B z = ( 1) + ( 1) ( 4) = 5 (b) (6 pts) Calculate the angle between A and B Solution: A B = A B cos θ A cos θ = B A B ) ( A B θ = cos 1 A B ( ) = cos ( 1) ( 1) 2 + ( 4) 2 ( ) 5 = cos = 73 (16 pts ) 2. A 3.0 box slides 5.0 m down a 20 ramp at a constant speed. (a) (4 pts) What was the total work done by all the forces acting on the box (consider the Work-Energy Theorem)? Solution: The Work-Energy Theorem says that the total work done by all forces is equal to the change in kinetic energy. Since the box is traveling at a constant speed, there is no change in the kinetic energy. Therefore, the total work done by all the forces acting on the box equals zero. (b) (4 pts) What was the work done by the Normal Force? Solution: The normal force is perpendicular to the direction the box is moving, θ = 90, so W N = F N r = F N r cos θ = 0 J 16 pts (c) (4 pts) What was the work done by the Gravitational Force? Solution: There are many ways to calculate this. For example, we know that the work done by gravity is always W g = m g h. If the box slides 5.0 m down a ramp at 20, h = (5 m) sin(20 ). 24 pts

2 AP Physics C/Work-Energy Mock Exam Page 2 of 6 Name: Answer Key Alternatively, we can use W g = F g r cos θ. If the ramp is at an incline of 20, the angle between the gravitational force and r is 70. Using the former method, we get W g = m g h = m g (h sin(20 )) = 50.3 J (d) (4 pts) What was the work done by the Frictional Force? Solution: Since friction, the normal force, and gravity are the only forces acting on the box, and we know that the total work done is zero, we can conclude that the work done by friction must be 50.3 J. Alternatively, you could calculate the work the long way (using the definition of work). To do this, you would first need to determine the magnitude of the frictional force using Newton s Second Law. Since the acceleration is zero, we know that the frictional force perfectly cancels the component of the gravitational force pulling the box down the ramp. Thus f = F g,x = m g sin θ. (16 pts ) 3. A 3.0 kg box lies on a frictionless surface. The box is attached to the end of a spring which has a spring constant of 1200 N. The box is pulled, stretching the spring by 15 cm. The box is then released from m rest. (a) (8 pts) How fast is the box moving when it passes through equilibrium (when the spring is no longer stretched)? Solution: Since there are no non-conservative forces in this part of the problem, we can find v f using conservation of energy: 16 pts E 0 = E f 1 2 k ( x)2 = 1 2 m v2 f vf 2 = k m ( x)2 k m x = 3 m s (b) (8 pts) Suppose a constant frictional force of 8.5 N is exerted on the box. How fast is the box traveling when it passes through equilibrium? Solution: In this case, we can not use conservation of energy, because friction is a non-conservative force. However, we can say that the work done by friction is equal to the change in the total energy of the system: 20 pts

3 AP Physics C/Work-Energy Mock Exam Page 3 of 6 Name: Answer Key W f = E f x = E f E ( 0 ) ( ) f x 1 1 = 2 m v2 f 2 k ( x)2 1 2 m v2 f = 1 2 k ( x)2 f x k( x) 2 2 f x m = 2.85 m s 0 pts

4 AP Physics C/Work-Energy Mock Exam Page 4 of 6 Name: Answer Key (12 pts ) 4. A 5 kg object is subject to a force which varies as a function of the object s position as shown in the figure below F (N) 12 pts x (m) (a) (4 pts) Determine the work done by this force when the object moves from x 0 = 0 m to x f = 6.0 m. Solution: Recall that when the force varies as the object moves, the work done by this force is given by: W = F d r. When looking at a graph of F vs. x, this integral is equal to the area under the graph. W = Area = 1 (10 N)(6 m) 2 = 30 J (b) (4 pts) Assuming the object starts from rest and that no other forces are exerted on the box, determine the velocity of the object at x f = 6.0 m. Solution: To find v f, we just use the Work-Energy Theorem: W = KE W = 1 2 m 1 v2 f 0 2 m v2 0 2 W m = 3.46 m s (c) (4 pts) Suppose the object started at x 0 with an initial speed of 2.5 m, what is the velocity of the s object at x f = 6.0 m? Solution: It is very tempting to say that v f will be 3.46 m greater than the initial velocity, however s this is not correct. The reason is that the kinetic energy is proportional to v 2, not v, and v0 2 + vf 2 is not the same thing as (v f + v 0 ) 2. To find v f, we need to go back to the Work-Energy Theorem: 12 pts

5 AP Physics C/Work-Energy Mock Exam Page 5 of 6 Name: Answer Key W = KE W = 1 2 m v2 f 1 2 m v2 0 v 2 f v 2 0 = 2 W m vf 2 = 2 W m + v2 0 2 W m + v2 0 = 4.27 m s 0 pts

6 AP Physics C/Work-Energy Mock Exam Page 6 of 6 Name: Answer Key (6 pts ) 5. An object s total potential energy is shown in the graph below. No non-conservative forces act on the object. U 6 pts (a) (3 pts) Identify the position of all stable equilibria. x (m) Solution: Recall that stable equilibria occur when the object is in a potential energy minimum. This occurs at x = ±3. (b) (3 pts) Identify the position of all unstable equilibria. Solution: Unstable equilibria occur when the object is in a potential energy maximum. This occurs at x = 0. 6 pts

Torque/Rotational Energy Mock Exam. Instructions: (105 points) Answer the following questions. SHOW ALL OF YOUR WORK.

AP Physics C Spring, 2017 Torque/Rotational Energy Mock Exam Name: Answer Key Mr. Leonard Instructions: (105 points) Answer the following questions. SHOW ALL OF YOUR WORK. (22 pts ) 1. Two masses are attached