The mass of the hanging mass is m and the mass of the cart is M.

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The mass of the hanging mass is m and the mass of the cart is M. You can assume that the floor over which the mass hangs has been chosen as the zero point of gravitational potential energy. If the hanging mass is at a height
above the floor, and has not yet been released, what type(s) of energy is/are present in the system? NOTE: You can ignore any energies that are not relevant (will not change) as the mass is held, released, and allowed to fall. What is the final energy of the system after the weight has traveled a distance
? Leave your answer in symbolic terms. Answer in terms of
,
,
, and/or
, and use
for the acceleration due to gravity.

Assume that energy is conserved. What is the final velocity of the cart, based on your expressions for initial and final energy? Leave your answer in symbolic terms. Answer in terms of
,
, and/or
, and use
for the acceleration due to gravity.
What is the acceleration of the cart-mass system to this point? Leave your answer in symbolic terms. Answer in terms of
,
, and/or
, and use
for the acceleration due to gravity. HINT: How are velocity, acceleration, and traveled distance related to each other? Look at your Lab02 results/derivations.

The Correct Answer and Explanation is:

Let’s analyze the system step-by-step and derive symbolic expressions for:

  • Initial energy
  • Final energy
  • Final velocity
  • Acceleration

System Overview

We have:

  • A cart of mass MMM
  • A hanging mass mmm
  • The hanging mass is at height hhh above the floor (initially), and it’s connected to the cart via a pulley system
  • Distance fallen by the mass: ddd
  • Gravity: ggg
  • Assumption: No friction, no energy loss — conservation of mechanical energy

1. Initial Energy

At the start:

  • The cart is stationary: no kinetic energy
  • The hanging mass is stationary but elevated: gravitational potential energy is present

Thus, total initial energy EiE_iEi​:Ei=mghE_i = mghEi​=mgh

2. Final Energy

After the hanging mass falls a vertical distance ddd:

  • The cart (mass MMM) and the hanging mass (mass mmm) both move with the same speed vvv
  • The cart is moving horizontally, the hanging mass is falling vertically
  • Final kinetic energy is:

Kf=12Mv2+12mv2K_f = \frac{1}{2}Mv^2 + \frac{1}{2}mv^2Kf​=21​Mv2+21​mv2

  • Final potential energy of hanging mass is:

Uf=mg(h−d)U_f = mg(h – d)Uf​=mg(h−d)

So total final energy EfE_fEf​:Ef=12Mv2+12mv2+mg(h−d)E_f = \frac{1}{2}Mv^2 + \frac{1}{2}mv^2 + mg(h – d)Ef​=21​Mv2+21​mv2+mg(h−d)

3. Use Energy Conservation

Ei=Ef⇒mgh=12Mv2+12mv2+mg(h−d)E_i = E_f \Rightarrow mgh = \frac{1}{2}Mv^2 + \frac{1}{2}mv^2 + mg(h – d)Ei​=Ef​⇒mgh=21​Mv2+21​mv2+mg(h−d)

Cancel mghmghmgh from both sides:0=12Mv2+12mv2−mgd0 = \frac{1}{2}Mv^2 + \frac{1}{2}mv^2 – mgd0=21​Mv2+21​mv2−mgd

Multiply through by 2:0=Mv2+mv2−2mgd⇒v2(M+m)=2mgd⇒v=2mgdM+m0 = Mv^2 + mv^2 – 2mgd \Rightarrow v^2(M + m) = 2mgd \Rightarrow v = \sqrt{\frac{2mgd}{M + m}}0=Mv2+mv2−2mgd⇒v2(M+m)=2mgd⇒v=M+m2mgd​​

4. Acceleration of the System

Use the kinematic relation:v2=2ad⇒a=v22dv^2 = 2ad \Rightarrow a = \frac{v^2}{2d}v2=2ad⇒a=2dv2​

Substitute v2=2mgdM+mv^2 = \frac{2mgd}{M + m}v2=M+m2mgd​:a=2mgd(M+m)⋅2d=mgM+ma = \frac{2mgd}{(M + m) \cdot 2d} = \frac{mg}{M + m}a=(M+m)⋅2d2mgd​=M+mmg​

Final Answers

  • Final velocity: v=2mgdM+mv = \sqrt{\frac{2mgd}{M + m}}v=M+m2mgd​​
  • Acceleration: a=mgM+ma = \frac{mg}{M + m}a=M+mmg​

Explanation

This problem involves analyzing the motion of a coupled system: a cart of mass MMM connected to a hanging mass mmm via a pulley. Initially, the cart and mass are at rest, with the hanging mass elevated at height hhh. The gravitational potential energy of the hanging mass is the only mechanical energy present, as nothing is moving. The system’s total energy is thus Ei=mghE_i = mghEi​=mgh.

When the mass is released and falls a distance ddd, it accelerates, pulling the cart horizontally. Both the cart and the mass move at the same speed vvv due to the inextensible rope. As the system moves, the gravitational potential energy of the hanging mass decreases, while kinetic energy increases for both masses. The final energy consists of the remaining potential energy of the hanging mass and the kinetic energy of both the cart and the mass.

Applying conservation of energy gives an expression relating vvv, ggg, mmm, MMM, and ddd. Solving the equation yields the final velocity:v=2mgdM+mv = \sqrt{\frac{2mgd}{M + m}}v=M+m2mgd​​

This shows that a greater mass mmm, a larger fall distance ddd, or a smaller cart mass MMM increases the final speed.

Next, using the kinematic relation v2=2adv^2 = 2adv2=2ad, we isolate aaa, the system’s acceleration, and substitute the earlier expression for v2v^2v2. This leads to:a=mgM+ma = \frac{mg}{M + m}a=M+mmg​

This is consistent with Newton’s second law and shows that acceleration depends on the net force (due to gravity on mass mmm) divided by the total mass of the system.

This analysis shows how energy conservation can yield deep insight into motion, without requiring force diagrams or direct Newtonian analysis.

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