Change in Kinetic Energy of block

[grouchy]

A 4.5 kg block is set into motion up an inclined plane with an initial speed of 8.9 m/s. The block comes to a rest after traveling 6 m along the plane, which is inclined at an angle of 24 to the horizontal. Determine the magnitude of the change in the block's kinetic energy.

I tried Delta K= Work of the Force
so..

Delta K = mgh but I got it wrong.

 

[rl.bhat]

Delta K = mgh
What is the value of h you have substituted in this equation?

 

[Doc Al]

I tried Delta K= Work of the Force
so..

Delta K = mgh but I got it wrong.

You don't need any such equation to answer the question. Instead: What's the initial KE? What's the final KE? What's the change?

 

[grouchy]

I did also try Delta K = 1/2mVf^2 - 1/2 mVi^2. The final is 0 since it is at rest and I got -178.222 for the initial. But that was also wrong

 

[grouchy]

Delta K = mgh
What is the value of h you have substituted in this equation?

for h I user 6sin24.

 

[Doc Al]

I did also try Delta K = 1/2mVf^2 - 1/2 mVi^2. The final is 0 since it is at rest and I got -178.222 for the initial. But that was also wrong

The initial KE = + 178.22 J. The change in KE is negative. But note that they ask for the magnitude of the change. (So drop the negative sign.)

 

[grouchy]

ok, I get that now. Magnitude is just the number and not the sign.

But for the next part it says that (Assumed: The fristional force is constant. Determine the magnitude of the frictional force exerted on the block.)

do I use, Wf = DeltaK + DeltaU?

 

[Doc Al]

do I use, Wf = DeltaK + DeltaU?

That will work.

 

[grouchy]

I don't get it, I'm getting it wrong. a previous question asked for the change in potential energy which I got to be 107.623. So I think I'm using the wrong sign or something...

I tried -178.223 + 107.623 and got -70.61N. I tried it with and without the negative sign and still got it wrong.

 

[Doc Al]

I tried -178.223 + 107.623 and got -70.61N. I tried it with and without the negative sign and still got it wrong.

That's the work done by friction, in Joules not Newtons. Now find the friction force.

 

[grouchy]

man..I suck at this physics stuff lol. I forgot that I found the W and not the force. Thx for your help though!