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AP Physics 1- 2.8 Spring Forces - Exam Style questions - FRQs- New Syllabus

Spring Forces AP  Physics 1 FRQ

Unit: 2. Force and Translational  Dynamics

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Question


A small block of mass \( m_0 \) is attached to the end of a spring of spring constant \( k_0 \) that is attached to a rod on a horizontal table. The rod is attached to a motor so that the rod can rotate at various speeds about its axis. When the rod is not rotating, the block is at rest and the spring is at its unstretched length \( L \), as shown. All frictional forces are negligible.
 
(a) At time \( t=t_1 \), the rod is spinning such that the block moves in a circular path with a constant tangential speed \( v_1 \) and the spring is stretched a distance \( d_1 \) from the spring’s unstretched length, as shown in Figure \( 1 \). At time \( t=t_2 \), the rod is spinning such that the block moves in a circular path with a constant tangential speed \( v_2 \) and the spring is stretched a distance \( d_2 \) from the spring’s unstretched length, where \( d_2 > d_1 \), as shown in Figure \( 2 \).
(i) On the following dots, which represent the block at the locations shown in Figure \( 1 \) and Figure \( 2 \), draw the force that is exerted on the block by the spring at times \( t=t_1 \) and \( t=t_2 \). The force must be represented by a distinct arrow starting on, and pointing away from, the dot.
Note: Draw the relative lengths of the vectors to reflect the relative magnitudes of the forces exerted by the spring at both times.
(ii) Referencing \( d_1 \) and \( d_2 \), describe your reasoning for drawing the arrows the length that you did in part \( \mathrm{(a)(i)} \).
(iii) Is the tangential speed \( v_1 \) of the block at time \( t=t_1 \) greater than, less than, or equal to the tangential speed \( v_2 \) of the block at time \( t=t_2 \) ?
_____ \( v_1 > v_2 \)      _____ \( v_1 < v_2 \)      _____ \( v_1 = v_2 \)
Justify your answer without using equations.
(b) Consider a scenario where the block travels in a circular path where the spring is stretched a distance \( d \) from its unstretched length \( L \).
(i) Determine an expression for the magnitude of the net force \( F_{\text{net}} \) exerted on the block. Express your answer in terms of \( m_0 \), \( k_0 \), \( L \), \( d \), and fundamental constants, as appropriate.
(ii) Derive an equation for the tangential speed \( v \) of the block. Express your answer in terms of \( m_0 \), \( k_0 \), \( L \), \( d \), and fundamental constants, as appropriate.
(c) Does your equation for the tangential speed \( v \) of the block from part \( \mathrm{(b)(ii)} \) agree with your reasoning from part \( \mathrm{(a)} \) ?
_____ Yes      _____ No
Explain your reasoning.

Most-appropriate topic codes (AP Physics 1):

• Topic \( 2.8 \) — Spring Forces (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)(i)} \))
• Topic \( 2.9 \) — Circular Motion (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)(ii)} \), Part \( \mathrm{(c)} \))
• Topic \( 2.5 \) — Newton’s Second Law (Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \))
• Topic \( 1.3 \) — Representing Motion (Part \( \mathrm{(a)(i)} \))
▶️ Answer/Explanation

(a)(i)
In both diagrams, the spring force on the block points horizontally toward the rod, which is to the right in the given figures.

So both arrows should point to the right. The arrow at \( t=t_2 \) should be longer than the arrow at \( t=t_1 \).

(a)(ii)
Because \( d_2 > d_1 \), the spring is stretched more at time \( t=t_2 \). A greater stretch means a greater spring force.

Therefore, the force arrow at \( t=t_2 \) must be longer than the force arrow at \( t=t_1 \). The two arrows point in the same direction, but the one for \( t=t_2 \) has larger magnitude.

(a)(iii)
\( \boxed{v_1 < v_2} \)

The spring force is the net inward force that causes the circular motion. Since the spring is stretched more at \( t=t_2 \), the spring force is greater at \( t=t_2 \). A greater inward net force corresponds to a greater centripetal acceleration, so the block must be moving faster at \( t=t_2 \).

Therefore, the tangential speed is larger when the spring is stretched farther.

(b)(i)
The only horizontal inward force on the block is the spring force, so the net force equals the spring force.

Using Hooke’s law,

\( F_{\text{net}} = F_s = k_0 d \)

Therefore, \( \boxed{F_{\text{net}} = k_0 d} \)

Note that \( L \) is the unstretched length, but the force depends on the amount of stretch, which is \( d \).

(b)(ii)
Start with Newton’s second law for circular motion:

\( \sum F = m a_c \)

The net inward force is the spring force:

\( k_0 d = \dfrac{m_0 v^2}{r} \)

The radius of the circular path is the total spring length,

\( r = L + d \)

so

\( k_0 d = \dfrac{m_0 v^2}{L+d} \)

Solving for \( v \),

\( v^2 = \dfrac{k_0 d (L+d)}{m_0} \)

\( \boxed{v = \sqrt{\dfrac{k_0 d (L+d)}{m_0}}} \)

This expression shows that the speed increases as the stretch \( d \) increases, because both \( d \) and \( L+d \) increase.

(c)
\( \boxed{\text{Yes}} \)

The equation from part \( \mathrm{(b)(ii)} \), \( v = \sqrt{\dfrac{k_0 d (L+d)}{m_0}} \), shows that the tangential speed increases when \( d \) increases.

Since \( d_2 > d_1 \), the equation predicts \( v_2 > v_1 \), which matches the reasoning in part \( \mathrm{(a)} \) that the block moves faster when the spring is stretched more.

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