Calculating Final Velocity in a System with Potential Energy and Friction

[closer]

The coefficient of friction (uk) between the 3.00 kg block (m1) and the surface in Figure P7.21 is 0.200. The system starts from rest. What is the speed of the 5.00 kg ball (m2) when it has fallen 1.20 m (y)?

Wnet = \DeltaKE
U_m2 - W_fk = .5m₂v_f² - .5m₂v_i²
(m₂g)(y) - (uk)(N)(x) = 5m₂v_f² - .5m₂v_i²
(m₂g)(y) - (uk)(m₁g)(x) = .5m₂v_f² - .5m₂v_i²

m1 = 3
m2 = 5
uk = .2
y = 1.2
vi = 0
vf = ?
x = ?

I'm trying to find vf. I don't know what x would be for the work done by friction. I'm sure I made some other mistakes as well. Thanks in advance.

 

[HallsofIvy]

Yes, you can use those formulas- why haven't you?

After the ball has fallen 1.5 m, its potential energy has decreased by its mass times g times the distance: that's the mgy in your second equation. Here m= 5.00, g= 9.81, and y= 1.5. Some of that energy went into friction (and warms the desk): that's the ukNx in your formula. You are told that uk= 0.200, N is the weight of the block, (3.00)(9.81, and x is the distance the block moved, also 1.5 since the rope connecting the block and ball doesn't change length. On the right of your second equation you have the total kinetic energy of the system. But you do not have vi and vf: both block and ball have 0 speed initially and the same speed finally. It should be (.4)(3.00+ 5.00)v² and it is v you want to solve for.

 

[closer]

(m₂g)(y) - (uk)(m₁g)(x) = .5m₂v_f² - .5m₂v_i²

(5)(9.8)(1.2) - (.2)(3)(9.8)(1.2) = (.5)(5)(v_f²) - 0
vf = 4.54

Final answer is incorrect. Any ideas?

Edit: Got it. As you just mentioned, I should have used mass of the system instead of just m2 on the right side.

vf = 3.59

Thanks.