# Homework Help: Tension of Rope, Friction, And Work Done

1. Nov 17, 2009

### Extremist223

A man drags a 25.0kg crate, by a rope, across the floor whose coefficient of kinetic friction is 0.250. The tension in the rope is 200.0 N at an angle of 30.0° to the vertical. What is the net work done on the box as it travels 10.0 m?

well I broke the tension force into the x and y components with, these equations.
cos60x200 = 100N
sin60x200=173.205N
also the Frictional force in the opposite direction of +x movement is fk=ukFN
fk= 0.25 x 25kg x 9.8m/s^2 = 61.25N

This is where I'm stuck, I thought I could subtract the x components and create a new triangle, find the new angle and determine the work done with W= Force x Distance cos theta. The answer I'm getting seems to be wrong.

Please answer this as soon as possible, I have a midterm tommorow and I'm struggling to teach myself. Thank you.

2. Nov 17, 2009

### Extremist223

oh I finally got it, I was forgetting that the y comp of tension was making the object lighter, so the answer is 820 Joules.

3. Nov 17, 2009

### PhanthomJay

OK,(100 N is the x component, and 173 N up is the y component)
The Normal force and the weight are not the same. Sum forces in y direction to find it.
rather, you should find the net force in the x direction
try using W_net = F_x(net) times distance. Is there any F_net in the y direction?