# Block sliding on top of cart

Homework Statement
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?

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phinds
Gold Member
2019 Award
The cart's rollers are frictionless ... What distance does the cart travel ... ?
Once the cart starts to move, what is going to stop it?

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

phinds
Gold Member
2019 Award
nothing stops the cart, but the question is asking how much distance has it traveled until the block on top stops sliding
Ah. Good point. I missed that.

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 $a = - \mu g$.

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?

stevendaryl
Staff Emeritus
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?
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:
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.)

TSny
Homework Helper
Gold Member
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.

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