Conservation of Energy AP Physics 1 FRQ

Conservation of Energy AP Physics 1 FRQ – Exam Style Questions etc.

Conservation of Energy AP Physics 1 FRQ

Unit: 3. Work , Energy and Power

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Question

In the figure above, a small object slides down a frictionless quarter-circular slide of radius R. If the object starts from rest at a height equal to 2R above a horizontal surface, find its horizontal
displacement, x, at the moment it strikes the surface.
(A) 2R
(B) \(\frac{5}{2}R\)
(C) 3R
(D) \(\frac{7}{2}R\)
(E) 4R

Answer/Explanation

Ans: C

The object’s initial velocity from the slide is horizontal, so \(v_{oy}=0\), which implies that \(\Delta = -\frac{1}{2}gt^2\). Since \(\Delta y = -R\),
\(-R = -\frac{1}{2} gt^2 \Rightarrow t = \sqrt{\frac{2R}{g}}\)
The (horizontal) speed with which the object leaves the slide is found from the conservation of energy: \(mgR = \frac{1}{2} mv^2\), which gives \(v=\sqrt{2gR}\). Therefore,
\(\Delta x = v_{ox} t = \sqrt{2g R} . \sqrt{\frac{2R}{g}} = 2R\)
Since the object travels a horizontal distance of 2R. from the
end of the slide, the total horizontal distance from the
object’s starting point is R + 2R = 3R.

Question

A pendulum consists of a small object of mass M and width d attached to a string of length L. In an experiment, the pendulum is pulled back so that the string makes an angle θ with the vertical and is released from rest. The object’s path is shown by the dashed arc, and its velocity is determined by measuring the time t it takes the object to pass through a photogate located at the lowest point in the path. The speed v is then calculated by dividing d by t. Given θ and L , the object’s kinetic and potential energies are calculated and compared. Which of the following sources of experimental error could cause the calculated kinetic energy of the object at the bottom of the path to be larger than the calculated value for the change in gravitational potential energy of the object-Earth system as the object moves from its release point to the lowest point in its path?

A The measurement of L is larger than the actual value.

B The measurement of θ is larger than the actual value.

C Air resistance is ignored.

D The measurement of d is larger than the actual value.

E The measurement of M is larger than the actual value.

Answer/Explanation

Ans:D

An increase in the distance d would increase the calculated kinetic energy; thus, the calculated kinetic energy would be larger than the change in potential energy.

Question

A object of mass 4.0kg that is being pushed or pulled by an ideal spring and the total mechanical energy of the object, and the potential energy of the object-spring system are shown in the graph provided. The object’s speed when it is at position \(−0.5m\) is most nearly

A \(2.45m/s\)

B \(3.54m/s\)

C \(4.36m/s\)

D \(19.0m/s\)

E \(37.5m/s\)

Answer/Explanation

Ans:C

Using the values from the graph and the equation for total mechanical energy, \(ME=K+U\)
                                                                                                                                  \( 50J=\frac{1}{2}mv^2+12J\)
                                                                                                                                 \(\frac{1}{2}(4)v^2=28J\)
                                                                                                                                    \( v=4.36 m/s\)

Question

The figure shows an object moving from point B to point A and the gravitational potential energy of the object-Earth system at different points along the trajectory. There are no other forces exerted on the object.

The magnitude of the force associated with the gravitational field is constant and has a value F . A particle is launched from point B with an initial velocity and reaches point A having gained \(U_0\) joules of kinetic energy. A resistive force field is now set up such that it is directed opposite the gravitational field with a force of constant magnitude \(\frac{1}{2}F\). A particle is again launched from point B. How much kinetic energy will the particle gain as it moves from point B to point A ?

A \(\frac{1}{4}U_0\)

B \(\frac{1}{2}U_0\)

C \(U_0\)

D \(\frac{3}{2}U_0\)

E \(2U_0\)

Answer/Explanation

Ans:B

The original force field increases the particle’s kinetic energy by \(U_0\) . According to the equation \(U=−∫Fdx\), since the resistive force has a magnitude of \(\frac{1}{2}F\), it will take away \(\frac{1}{2}U_0\) of the kinetic energy of the particle; thus, the particle will gain \(U_0−\frac{1}{2}U_0=\frac{1}{2}U_0\).

Question

A student pushes a box across a rough horizontal floor. If the amount of work done by the student on the box is 100 J and the amount of energy dissipated by friction is 40 J, what is the change in kinetic energy of the box?

A 0 J

B 40 J

C 60 J

D 100 J

E 140 J

Answer/Explanation

Ans:C

Conservation of Energy AP Physics 1 FRQ