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AP Physics 1- 3.3 Potential Energy - Exam Style questions - FRQs- New Syllabus

Potential Energy AP  Physics 1 FRQ

Unit: 3. Work , Energy and Power

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Question

A block of mass \( M \) is released from rest at position \( x=0 \) near the top of a ramp. The ramp makes an angle of \( \theta \) with the horizontal. The block slides down the ramp with negligible friction. At \( x=8D \) the block makes contact with an uncompressed spring with spring constant \( k \). The spring is then compressed and the block momentarily comes to rest at \( x=12D \). Figure \( 1 \) shows the instants when the block is at \( x=0 \), \( x=6D \), and \( x=10D \), respectively.
A. Figure \( 4 \) shows an energy bar chart that represents the kinetic energy \( K \) of the block, the gravitational potential energy \( U_g \) of the block-spring-Earth system, and the spring potential energy \( U_s \) of the block-spring-Earth system at the instant that the block is at \( x=10D \). The gravitational potential energy \( U_g \) of the block-spring-Earth system is defined to be zero when the block momentarily comes to rest at \( x=12D \).
Draw shaded bars that represent \( K \), \( U_g \), and \( U_s \) to complete the energy bar charts in Figure \( 2 \) and Figure \( 3 \) for when the block is released from rest at \( x=0 \) and for when the block is at \( x=6D \), respectively.
• Shaded bars should start at the dashed line that represents zero energy.
• Represent any energy that is equal to zero with a distinct line on the zero-energy line.
• The relative heights of each shaded bar should reflect the magnitude of the respective energy consistent with the scale used in Figure \( 4 \).
B. Figure \( 5 \) shows the block at \( x=0 \) when the block is released from rest and the block at \( x=12D \) when the block momentarily comes to rest against the compressed spring.
Starting with conservation of energy, derive an equation for the spring constant \( k \). Express your answer in terms of \( M \), \( \theta \), \( D \), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from reference information.
C. Figure \( 6 \) shows a graph of the energy of the system as a function of the position of the block from \( x=8D \) to \( x=12D \). The spring potential energy \( U_s \) of the block-spring-Earth system is shown on the graph.
On the axes shown in Figure \( 6 \), do the following.
(i) Sketch and label a line or curve that represents the total mechanical energy \( E \) for the block-spring-Earth system as a function of the position of the block from \( x=8D \) to \( x=12D \).
(ii) Sketch and label a line or curve that represents the gravitational potential energy \( U_g \) for the block-spring-Earth system as a function of the position of the block from \( x=8D \) to \( x=12D \).
D. Indicate whether the speed \( v_{9D} \) of the block at \( x=9D \) is greater than, less than, or equal to the speed \( v_{8D} \) of the block at \( x=8D \).
_____ \( v_{9D} > v_{8D} \)
_____ \( v_{9D} < v_{8D} \)
_____ \( v_{9D} = v_{8D} \)
Justify how your response is consistent with the energy lines or curves you drew in Figure \( 6 \) in part \( \mathrm{C} \).

Most-appropriate topic codes (AP Physics 1):

• Topic \( 3.1 \) — Translational Kinetic Energy (Part \( \mathrm{A} \), Part \( \mathrm{D} \))
• Topic \( 3.3 \) — Potential Energy (Part \( \mathrm{A} \), Part \( \mathrm{B} \), Part \( \mathrm{C} \))
• Topic \( 3.4 \) — Conservation of Energy (Part \( \mathrm{A} \), Part \( \mathrm{B} \), Part \( \mathrm{C} \), Part \( \mathrm{D} \))
▶️ Answer/Explanation

A
For \( x=0 \): the block is released from rest, so \( K=0 \), \( U_s=0 \), and all the mechanical energy is gravitational potential energy.

Therefore, in Figure \( 2 \):
\( K=0 \)
\( U_g=12E_0 \)
\( U_s=0 \)

For \( x=6D \): the block has not yet touched the spring, so \( U_s=0 \). Since total mechanical energy is conserved and still equals \( 12E_0 \), the remaining energy is split between \( K \) and \( U_g \).

Therefore, in Figure \( 3 \):
\( K=6E_0 \)
\( U_g=6E_0 \)
\( U_s=0 \)

This is consistent with Figure \( 4 \), where at \( x=10D \) the energies add to \( 12E_0 \):
\( 7E_0 + 2E_0 + 3E_0 = 12E_0 \).

B
Start with conservation of energy:

\( E_i = E_f \)

At \( x=0 \), the block is released from rest, so all the energy is gravitational potential energy. At \( x=12D \), the block is momentarily at rest, so all the energy is spring potential energy.

Thus,

\( U_g = U_s \)

The vertical drop from \( x=0 \) to \( x=12D \) is

\( \Delta y = 12D\sin\theta \)

The spring is first contacted at \( x=8D \), so the compression at \( x=12D \) is

\( \Delta x = 12D-8D = 4D \)

Therefore,

\( Mg(12D\sin\theta)=\dfrac{1}{2}k(4D)^2 \)

\( 12MgD\sin\theta = 8kD^2 \)

Solving for \( k \),

\( k=\dfrac{12MgD\sin\theta}{8D^2} \)

\( k=\dfrac{3Mg\sin\theta}{2D} \)

Therefore, \( \boxed{k=\dfrac{3Mg\sin\theta}{2D}} \)

C(i)
The total mechanical energy \( E \) is constant from \( x=8D \) to \( x=12D \), so the graph should be a horizontal line at \( 12E_0 \).

Even though \( U_s \), \( U_g \), and \( K \) individually change, their sum stays the same because friction is negligible.

C(ii)
The gravitational potential energy \( U_g \) decreases linearly as the block moves down the ramp from \( x=8D \) to \( x=12D \).

So the graph should be a straight line decreasing from \( (8D,4E_0) \) to \( (12D,0) \).

The decrease is linear because height decreases linearly with distance traveled along a straight ramp.

D
\( \boxed{v_{9D} > v_{8D}} \)

At \( x=8D \), the spring is just uncompressed, so \( U_s=0 \). From the graph in part \( \mathrm{C} \), the total energy is \( 12E_0 \) and \( U_g=4E_0 \), so the kinetic energy at \( x=8D \) is

\( K_{8D}=12E_0-4E_0-0=8E_0 \)

At \( x=9D \), the spring potential energy is still less than \( E_0 \) while the gravitational potential energy is \( 3E_0 \). Therefore the total of \( U_g+U_s \) at \( x=9D \) is less than \( 4E_0 \), which means the kinetic energy at \( x=9D \) is greater than \( 8E_0 \).

Since the mass is the same at both positions and \( K=\dfrac{1}{2}Mv^2 \), greater kinetic energy means greater speed. Therefore, \( v_{9D} > v_{8D} \).

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