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Potential Energy AP  Physics 1 FRQ

Potential Energy AP  Physics 1 FRQ – Exam Style Questions etc.

Potential Energy AP  Physics 1 FRQ

Unit: 3. Work , Energy and Power

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Exam Style Practice Questions, Potential Energy AP  Physics 1 FRQ

Question

The variable x represents the position of particle A in a two-particle system. Particle B remains at rest. The graph above shows potential energy U of the system as a function of x .

The x-component of the force on particle A when it is at x = 0.15 m is most nearly
(A) –20 N
(B) –2.0 N
(C) –1.0 N
(D) 2.0 N
(E) 20 N

Answer/Explanation

Ans:E

Question

An object of mass m is constrained to move along the x-axis only. It experiences a potential energy function \(U=\frac{1}{2}kx^{2}-bx\), where U is in joules and x is in meters. Which of the following represents the net force F on the object?
(A) \(b-kx\)
(B) \(kx-b\)
(C) \(\frac{kx^{3}}{6}-bx^{2}\)
(D) \(bx^{2}-\frac{kx^{3}}{6}\)
(E) \(kx+b\)

Answer/Explanation

Ans:

A—Take the negative derivative of a potential energy function to get force. Here, the force is –kx + b.

Question

                                       

Two hikers start from the top of the same hill but take different paths to the bottom. Hiker A weighs more than hiker B. The hikers take the paths shown in the figure. Hiker B takes a longer time to descend than hiker A. Which of the following is a correct statement about the change in gravitational potential energy \(ΔUA\) for the Earth-hiker A system and the change in gravitational potential energy \(ΔUB\) for the Earth-hiker B system?

A \(ΔUA=ΔUB\) , because the height descended for both is the same.

B \(ΔUA<ΔUB\), because the distance traveled along the path for hiker B is greater.

C \(ΔUA<ΔUB\), because the time required for hiker B to descend the hill is longer.

D \(ΔUA>ΔUB\), because the time required for hiker A to descend the hill is shorter.

E \(ΔUA>ΔUB\), because the gravitational force exerted on hiker A is greater.

Answer/Explanation

Ans:E

The change in gravitational potential energy depends on both the height descended and the weight of the hiker. Both hikers descend the same height. Therefore, since hiker A weighs more, the Earth-hiker A system has a greater change in gravitational potential energy.

Question

An object of mass M moves in one dimension along the x-axis. A conservative force \(F(x)\) is exerted on the object. The potential energy \(U(x)\) associated with this force as a function of position x is shown in graph 1. A student used the potential energy graph to construct the graph of \(F(x)\) as a function of x shown in graph 2. Are these graphs consistent with one another, and if not, what is the error?

                                                             

A For \(0<x<2m\) and \(3m<x<4m\), the graph of \(F(x)\) should be curved, not constant.

B For \(0<x<2m\) , \(F(x)\) should be negative, and for \(3m<x<4m\), \(F(x)\) should be positive.

C For \(0<x<2m\) , \(F(x)\) should be negative because the slope of \(U(x)\) is negative.

D For \(3m<x<4m\) , \(F(x)\) should be positive because \(U(x)\) is positive.

E The two graphs are consistent with each other.

Answer/Explanation

Ans:E

According to the equation \(F=\frac{−dU}{dx}\) , the force is the negative of the slope of the potential energy graph. All sections of the force graph have values that are the negative of the slope of the potential energy graph.

Question

A certain nonlinear spring has a force function given by \(F=−ax^2−b\), where x is the displacement of the spring from equilibrium, \(a=3.0N/m^2\), and b=4.0N. The change in elastic potential energy of the spring as it is stretched from x=0m to x=2.0m is

A \(16J\)

B \(12 J\)

C \(−8  J\)

D \(−12J\)

E \(−16J\)

Answer/Explanation

Ans:A

Using the equation \(U=−∫Fdx\) yields \(\Delta U=\int_{x=0}^{x=2}\left ( -ax^2-b \right )dx=\left [ -\left ( -\frac{1}{3}ax^3-bx \right ) \right ]_{x=0}^{x=2}\)
                                                                         \(\Delta U=\left [ \frac{1}{3} ax^3=bx\right ]_{x=0}^{x=2}=\left ( \frac{1}{3}(3) (2)^3+(4)(2)\right )-(0)=16J\)
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