# Block sliding on top of cart

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1. Dec 18, 2016

### Raj Kishore

The problem statement, all variables and given/known data
A block of mass m =0.75 kg is set on a large cart of mass M = 6.5 kg. The cart's rollers are frictionless, but there is a coefficient of kinetic friction of 0.36 between the cart's surface and the block. The cart is at rest when the block is given an initial speed of 5.2 m/s. The block slides across the cart's surface, but the sliding stops before the block reaches the other end of the cart.

What distance does the cart travel along the floor from the time the block is set in motion to the time sliding ceases?

The attempt at a solution
What I did to solve the problem is I first used conservation of momentum to find the final velocity of the cart.
I did mv0 = (m+M)vf .
Then I found acceleration by doing a = μmg/M.
Then I used those values and used v2 = v02 +2ax and solved for x, and got .35 meters, but it is wrong. Can someone help?

2. Dec 18, 2016

### phinds

Once the cart starts to move, what is going to stop it?

3. Dec 18, 2016

### Raj Kishore

nothing stops the cart, but the question is asking how much distance has it traveled until the block on top stops sliding

4. Dec 18, 2016

### phinds

Ah. Good point. I missed that.

5. Dec 18, 2016

### stevendaryl

Staff Emeritus
Your expression for $a$ is wrong. The force of friction is $F_f = - \mu mg$, where $m$ is the mass of the block. So the relevant acceleration is $a = - \mu g$.

6. Dec 18, 2016

### Raj Kishore

Isn't that the acceleration of the block? The acceleration of the cart would be Ma = μmg, so a = μmg/M... right? Or am I wrong?

7. Dec 18, 2016

### stevendaryl

Staff Emeritus
You're right--that's how far the block slides on the cart before coming to rest, which isn't what the problem asked for.

Last edited: Dec 18, 2016
8. Dec 18, 2016

### stevendaryl

Staff Emeritus
So in terms of the cart, we have the initial velocity is zero, and the final velocity is $v$, so we use your formula:

$v^2 = v_0^2 + 2 a x$

with $v_0 = 0$ and $a = \frac{\mu m g}{M}$ (as you said). (the acceleration of the cart is positive.)

9. Dec 18, 2016

### TSny

Raj, your work looks correct to me. With g = 9.8 m/s2, I get the distance to be 0.355 m = 0.36 m.

10. Dec 18, 2016

### Raj Kishore

Ok thanks guys, I talked to my teacher and he said he made a calculation error.