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Deathslayer7
2010-02-23, 10:43 PM
If a force F=200N is applied to the 28kg cart, determine the time for the 20kg block A to move on the cart 1.5 m. The coefficients of static and kinetic friction between the block and the cart are .3 and .25. Both the cart and the block start from rest.

Meaning a block is on top of a cart that is being pulled by a 200N force to the right.

Need someone to check my calculations and see where i went wrong.

The cart:

48kg*9.8= 470.4N (down). Normal force going up. To the right 200N.

Thus 470.4=normal force.

200=m*a(x)
a(x)=4.167 m/s^2

from there the block.

Forces are 4.167*20kg= 83.333 N to the right. Gravity down (9.8*20)= 196 N Friction to the left = .25*196= 49 N. And normal force up.

Normal force=196.
F=ma
so we get 83.333- 49 = 20a(x)

a(x)=1.72

integrate twice to get postion

1.5=0.858*t^2

t= 1.32s

should be somewhere around 1s.

help?

Miklus
2010-02-24, 05:04 PM
Ok...

The block pushes down on the cart with force 20kg * 9.8m/s2 = 196N

The friction force between the cart and block is 0.25 * 196N = 49N

So far so good...

The acceleration of the cart is the total force divided by the mass:
The total force the the 200N minus the friction from the block 200N - 49N = 151N. The friction from the lagging block slows the cart down.

Acc_cart = 151N/28kg = 5.39m/s2. The cart accelerates to the right.

The acceleration of the block is caused by the friction only:

A_block = 49N/20Kg = 2.45m/s2. This ALSO to the right, but not as fast as the cart.

The difference in acceleration, that is, the acceleration of the block in the carts referece frame is 5.39m/s2 - 2.45m/s2 = 2.94m/s2.

You got the last part right:

1.5m = 1/2 * 2.94m/s2 * t2

Solve for t

t = square root ( 2 * 1.5m / 2.94m/s2) = 1.01s

Emlyn
2010-02-24, 05:29 PM
Ahh physics, we had such fun times together. You trying to drown me and I not wanting to.

As for your problem, it looks right. Of course this is coming from someone who once got an imaginary velocity. :smallfrown: