Home / AP Physics 2: 2.4 Thermodynamics and Free-Body Diagrams – Exam Style questions with Answer- MCQ

AP Physics 2: 2.4 Thermodynamics and Free-Body Diagrams – Exam Style questions with Answer- MCQ

Question

                        

In the system shown, a movable piston is initially at rest inside a cylindrical vacuum chamber. The piston has area \(0.050m^2\) and mass 0.50kg . An explosion just above the piston produces a pressure of \(60,000Pa\) on the piston, causing it to rapidly accelerate downward . The free-body diagram of the piston just after the explosion is shown. What is the magnitude of the piston’s acceleration at this instant?
A \(120,000m/s^2\)

B \(30,000m/s^2\)

C \(6010m/s^2\)

D \(6000m/s^2\)

▶️Answer/Explanation

Ans: C

 Both the force of the explosion and the weight of the piston will add to the acceleration. To calculate the force exerted on the piston by the explosion, the relationship between pressure and force is used: \(F=PA=(60,000Pa)(0.050m2)=3000N\) . Applying Newton’s second law allows the acceleration to be determined: \(F_{explosion}=mg+ma\), \(a=\frac{F_{explosion}}{m}+g=\frac{300N}{0.50kg}+10m/s^2=6010m/s^2\).

Question

                     

An ideal gas is sealed in a cylindrical container with an open top and a moveable piston, as shown in the figure. The piston has a mass of 10kg and a radius of 10cm. At the instant shown, the pressure of the gas in the cylinder is \(105,000Pa\). The free-body diagram shows the forces acting on the piston at this instant, which are the force \(F_{Gas}\) exerted by the ideal gas in the container, the force \(F_{Air}\) exerted by the air above the piston, and the force \(F_g\) due to gravity. The magnitude of the acceleration of the piston is most nearly

A \(5.7 m/s^2\)

B \(10 m/s^2\)

C \(15.7 m/s^2\)

D \(26 m/s^2\)

▶️Answer/Explanation

Ans:A

Applying Newton’s second law to the three forces in the free-body diagram and using consistent signs for the directions allows one to solve for the acceleration: \(ma=F_{Gas}−F_{Air}−F_g\)
, \(a=\frac{F_{Gas}−F_{Air}−F_g}{m} =\frac{P_{Gas}A−P_{Air}A−F_g}{m}. Substituting the given values: a=\frac{(1.05×10^5Pa)(π0.1^2)−(1.0×10^5Pa)(π0.1^2)−(10kg)(10m/s^2)}{10kg} =5.7m/s^2\).

Question

A block with mass \(m_b\) sits at rest on a movable piston with mass \(m_p\) that is fitted in the top of a cylindrical tank filled with a gas. When the block is removed, the piston accelerates upward. Free-body diagrams for the piston before and after the block is removed are shown above. Which of the following gives the correct expression for the acceleration of the piston at the instant after the block is removed?

A \(\frac{m_b}{m_p}g\)

B \(g\)

C \((\frac{2m_p}{m_b}+1)g\)

D \((2+\frac{m_b}{m_p})g\)

▶️Answer/Explanation

Ans: A

 Using Newton’s second law before the block is removed gives \(∑F_{before}=F_{gas}−(W_{piston}+W_{block})=0\)
, so \(F_{gas}=(m_p+m_b)g\). Using Newton’s second law after the block is removed gives \(∑F_{after}=F_{gas}−W_{piston}=mp_a\), so \(a=\frac{F_{gas}−m_pg}{m_p}\). Substituting for \(F_{gas}\) gives \(a=\frac{(m_p+m_b)g−m_pg}{m_p}=\frac{m_b}{m_p}g\).

Question

The temperature of an ideal gas is held constant as its volume is slowly decreased. Which of the following claims correctly describes the cause of the increase in the force exerted by the gas on the walls of the container?

A The gas molecules strike the walls more frequently.

B The gas molecules have a greater change in velocity during collisions, which will cause a greater force.

C The interaction time of the gas molecules with the walls will be smaller, which will lead to a greater force.

D The gas molecules collide with each other more frequently, which increases the average kinetic energy of the molecules.

▶️Answer/Explanation

Ans:A

With a smaller volume for the molecules, there will be more collisions among the molecules and between the molecules and the walls of the container. With each collision with the container there will be a force between molecules and the container; more collisions will lead to a greater force, and, as a result, an increase in pressure.

Question

            

The graph shows pressure as a function of volume for a sample of an ideal gas that is confined in a cylinder. A moveable piston sits on top of the gas, and the piston can move vertically with negligible friction. The weight of the piston is significantly less than the force exerted by the gas on the piston during the process shown. Which of the following is true of the acceleration of the piston during the process?

A The acceleration is zero because the net force on the piston is zero.

B The acceleration is decreasing because the pressure decreases as the volume increases.

C The acceleration is increasing because the net force acting on the piston increases as the volume increases.

D The acceleration stays the same since the net force on the piston stays the same.

▶️Answer/Explanation

Ans:D

 The pressure stays the same throughout the process, so the force the gas exerts on the piston is constant. This means a constant net force is exerted on the piston, which produces a constant acceleration.

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