How Much Work Does Kinetic Friction Do on a Sliding Block?

[TheFlemster]

Homework Statement

A 5.72 kg block is released from rest on an inclined surface. The incline makes an angle of 25.0 degrees relative to horizontal. The block is known to be traveling at a speed of 1.33 m/s after sliding a distance of 2.60 m along the incline. How much work was done on the block by kinetic friction?

2. Homework Equations
W = F x d
W = F x d x cos()
K₁ + U₁ + W_other = K₂ + U₂

The Attempt at a Solution

I used all three equations and did not get the correct answer

 

[gneill]

I used all three equations and did not get the correct answer

Show us what you tried.

 

[TheFlemster]

 

[TheFlemster]

the first pic are the possible answer choices

 

[gneill]

You seem to be confusing force directions and motion directions in some cases, and using incorrect distances in others. Perhaps you need to create a sketch of the scenario with the various distances, forces, and angles indicated. A free body diagram would help.

In your second relevant equation, W = f d cos(θ), what does the angle θ represent?

I'd suggest that you calculate the work done by gravity (use conservation of energy and the work-energy theorem to make it simple) and the final kinetic energy as separate items to begin with.

 

[Cool4Kat]

I think your mistake is that the height is not 2.6 m, the length of the wedge is 2.6 meters. So the potential energy is mgh but you have the wrong h! (However, you can use trigonometry to find the height given the angle and the length of the wedge)

Now if you want to use the Work = Force*distance for this problem, then you need to find the force of friction. How do you do that? Well, try drawing a FBD and using sum of forces down the ramp = ma (note, you also need to find the acceleration of the block, but you have the distance and the change in speed so look at your equations of motion). This method is more work but should give you the same answer!

Hope that helps (but not too much) :)

 

[TheFlemster]

Thanks to both of yall. I found the height with trig and recalculated and got the correct answer.