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?

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?