# PLEASE see if I solved this correctly.

1. Oct 27, 2009

1. The problem statement, all variables and given/known data

A tractor exters a force of 1200N to pull a fallen tree to pull a fallen tree along the ground. If the tree has a mass of 320KG and the coefficient of sliding friction is 0.35, find
a. the force of sliding friction
b. the netforce on the tree
c. the acceleration of the tree

Note:In case of sliding friction, if the force applied to a body is greater than the force needed to overcome friciton, there is acceleration.

Two children are pulled on a sled over snow-covered ground. The sled which is initially at rest, is pulled by a rope that makes an angle of 40 degrees with the horizontal. The children have combined mass of 45 kg and the sled has a mass of 5kg. The coefficient of kinetic and static friction are U(k)=0.15 and U(s)= 0.2 . Find the frictional force exerted by the grounf on the sled and the acceleration of the children and sled, starting from rest, if the tension in the rope is a)100N b)140N

2. Relevant equations

Forces in the x direction is Fx: 1200 - friction = ma

3. The attempt at a solution
I think the force of sliding friction would be

1097.6

Acceleration of tree is .32m/s

force of sliding friction is the coefficient * Normal force

You can find normal force because Fy: N - Weight of tree = 0 That is the forces in the y direction.
N = weight of the tree and weight of the tree is mass times gravity.

N* .35 = 1097.6

Forces in the x direction is Fx: 1200 - friction = ma
1200-1097.6 = 320a
a = (1200-1097.6)/320
a = .32

I hope that's right >_>

2. Oct 27, 2009

Looks good.