Work done by a rope pulling a cart at 28.5°

[bossalinie786]

this is one of my homework problems, i have attempted it twice, and the website i submit my homework to tells me it is wrong, here is the question,

A cart loaded with bricks has a total mass of 16.9 kg and is pulled at constant speed by a rope. The rope is inclined at 28.5 degrees above the horizontal and the cart moves 22.2 m on a horizontal surface. The coefficient of kinetic friction between ground and
cart is 0793.
The acceleration of gravity is 9.8 m/s2 :
How much work is done on the cart by the
rope? Answer in units of J.


since the rope is pulling is cart, and the force and the direction of the work are both going in the direction, then the angle(theta) would be zero? therefore only the x component of the force is needed, i have calculated the Friction Force which to my understand is the same as the force of the weight pulled. then i simply plug it into the equation W=F*d and I'm not getting it right, can anyone help, it would be appreciated

 

[cookiemonster]

Care to show some of your work so far?

cookiemonster

 

[gameguru]

The force of the rope and the work done by the rope are not pointing the same direction. According to the problem, the rope is inclined 28.5 degrees above the horizon and therefore has both an x and y component. The x component of the rope's force (cosine theta) is what points in the same direction as the work. W = F(cos theta)s

 

[turin]

Originally posted by bossalinie786
The rope is inclined at 28.5 degrees above the horizontal ...
...
... the Friction Force ... to my understand is the same as the force of the weight pulled.

Or is it?