How can I solve this tricky problem involving work and kinetic energy?

[kappcity06]

this proplem is very tricky and i was wondering if anyone could help me.

A worker pushed a 27 kg block 10.4 m along a level floor at constant speed with a force directed 25° below the horizontal.
(a) If the coefficient of kinetic friction is 0.20, how much work was done by the worker's force?
?J
(b) What was the increase in thermal energy of the block-floor system? ?J

I decided to try using two equations. f=mewN/cos25 AND F=MG-n/SIN 25 . I equated them to fine part a but that answer did not work. any help

 

[sdekivit]

this proplem is very tricky and i was wondering if anyone could help me.

A worker pushed a 27 kg block 10.4 m along a level floor at constant speed with a force directed 25° below the horizontal.
(a) If the coefficient of kinetic friction is 0.20, how much work was done by the worker's force?
?J
(b) What was the increase in thermal energy of the block-floor system? ?J

I decided to try using two equations. f=mewN/cos25 AND F=MG-n/SIN 25 . I equated them to fine part a but that answer did not work. any help

What is the energy balance in this problem ?

 

[Doc Al]

I decided to try using two equations. f=mewN/cos25 AND F=MG-n/SIN 25 .

Is "f" meant to be the friction? If so, that first equation is incorrect.
Is "F" meant to be the worker's force? If so, that second equation has an incorrect sign.

 

[kappcity06]

ok i got two equations sumation of F_x=mgcos30-Force of kentic friction and f=m*mew*n cos 30.

 

[Doc Al]

...and f=m*mew*n cos 30.

Kinetic friction is simply:
f = \mu N

 

[kappcity06]

w=fdcos(theata) right

 

[kappcity06]

anyone else know something that can be done with this problem

 

[Doc Al]

It's certainly true that the work done by the force F is:
W = Fd \cos\theta

 

[kappcity06]

anyone know something that I should do.

 

[Doc Al]

Why don't you solve for F or F\cos\theta so you can calculate the work done? Hint: The block is moving at a constant speed.

 

[kappcity06]

i have already found the normal force

 

[kappcity06]

i used two equations F=u(k)N/cos(25) and F=(mg-N)/sin25

I solved them to get a normal force of 242.028.

I then subsitied the first equation F=u(k)N/cos(25) in for the F in
w=fdcos25. I found 503 and that was wrong. I have no idea what i did wrong.

 

[Doc Al]

i used two equations F=u(k)N/cos(25)

This one looks OK.

and F=(mg-N)/sin25

This one does not.

 

[kappcity06]

what should i change. should the N be +

 

[Doc Al]

Yes. But you should know why it's wrong and needs to be changed.

 

[kappcity06]

so it should me (mg+n)/sin25 because the normal force pushes off the ground

 

[kappcity06]

ok i got part a correct but i do not kow how to do part b. for a i got 603.8

 

[Doc Al]

What happens to the work done by the worker? Where does the energy go?

 

[kappcity06]

it turns into heat

 

[Doc Al]

So what's the answer to part b?

 

[kappcity06]

mg=603.8+E(of friction)

 

[kappcity06]

i'm still not sure how to approach part b. can anyone help

 

[Hootenanny]

i'm still not sure how to approach part b. can anyone help

HINT: Work done be frictional force is equal the the thermal energy supplied to the system.