Two carts, C and D, are initially at rest on a horizontal frictionless table. A constant force of magnitude F is exerted on each cart as it travels between two marks on the table. (Mark 1 = Starting; Mark 2 = Finish) Cart D has a greater mass than cart C. (Mc < Md)
Is the statement below true? Explain.
"Since the same force is exerted on both carts, the cart with the smaller mass will move quickly, while the cart with the larger mass will move slowly. Since the velocity is squared to get the kinetic energy but mass isn't, the cart with the bigger speed must have more kineric energy."
This boggles me. I'm getting mixed answers...help, please? I know that the first part of the statement is true (Cart c moves faster than cart d). However, when I just use KE = (1/2)mv^2 for both carts, the statement is true, but when I use W net ext = Δ KE, I get that they're the same. So, which one is it?
Answers & Comments
Verified answer
The statement, as written, is false. The speed of the smaller cart is greater, but both carts have the same kinetic energy. There are a couple ways to verify this.
One is to use the work-energy theorem: the change in energy is equal to the work done on each cart. The work done is equal to the force F times the displacement, which is the same for both carts (both move from mark 1 to mark 2). Since the same force F is applied, the total energy of both carts increases by the same amount. Since the surface is horizontal, there is no change in potential energy for either cart, so the KE of both carts increases by the same amount.
You can also solve it explicitly: if the mass M is accelerated by force F, to get total acceleration you set F = Ma, so a = F/M. Then, you can use the kinematic equation v² = v₀² + 2aΔx. Since KE is 1/2mv², we can multiply both sides by M/2 to get:
1/2Mv² = 1/2Mv₀² + (1/2M)(2a)Δx
KE = KE₀ + aMΔx
KE = KE₀ + (F/M)MΔx
KE = KE₀ + FΔx
Cool...we just proved the work-energy theorem. Also, notice that mass ended up cancelling out; thus, it doesn't matter which mass we push. As long as F and Δx are the same, the change in KE is the same.