Work and Energy - angular force with friction

[estafusis]

Homework Statement

A cart loaded with bricks has a total mass of 13.8 kg and is pulled at constant speed by a rope. The rope is inclined at 20.8° above the horizontal and the cart moves 30.4 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.7 . The acceleration of gravity is 9.8 m/s$^{2}$ . How much work is done on the cart by the rope?

Homework Equations

I'm clearly missing something I need because I can't really get close to a solution.

F$_{friction}$ = $\mu_{k}$ * F$_{normal}$
ω = F$_{rope}$ * Δx * COSθ

The Attempt at a Solution

I've gotten this far, but there must be something I'm missing because I still can't get any kind of solution out of it.
F$_{g}$ = 135.2N
F$_{rope}$x = F$_{friction}$
F$_{n}$ = F$_{g}$ - F$_{rope}$y
F$_{rope}$x = F$_{rope}$ * COSθ
F$_{rope}$y = F$_{rope}$ * SINθ

ω = F$_{rope}$ * 30.4 * COS(20.8)
F$_{rope}$ = ?

I've got about 15 more problems just like this to be done tonight so quick responses would be awesome and I REALLY hope its just some small equation or equality that I'm missing.

 

[anathema]

Hello there,

From the given information, we can use the formula for work, W = F * d * cosθ, where F is the force applied, d is the distance traveled, and θ is the angle between the force and the direction of motion.

In this case, the force applied is the tension in the rope, F_rope. The distance traveled is 30.4 m on a horizontal floor, and the angle between the force and the direction of motion is 20.8°.

So, W = F_rope * 30.4 * cos20.8°

Now, we need to find the tension in the rope, F_rope. To do this, we can use the equation for the sum of forces in the x-direction, which is given by:

ΣF_x = ma_x

In this case, the only force acting in the x-direction is the force of friction, F_friction. So, we can write:

F_friction = μ_k * F_normal

And since we know that F_normal = F_g - F_rope * sin20.8°, we can substitute this into the equation above to get:

F_friction = μ_k * (F_g - F_rope * sin20.8°)

Now, we can substitute this into the equation for ΣF_x and solve for F_rope:

F_friction = ma_x

μ_k * (F_g - F_rope * sin20.8°) = ma_x

μ_k * (135.2 - F_rope * sin20.8°) = 0

Solving for F_rope, we get:

F_rope = 135.2 / sin20.8°

Substituting this into the equation for work, we get:

W = (135.2 / sin20.8°) * 30.4 * cos20.8°

Solving for W, we get:

W = 255.3 J

So, the work done on the cart by the rope is 255.3 J.

I hope this helps! Let me know if you have any further questions or if you need any clarification.