A block of mass 1.00 sits on an inclined plane as shown. (Intro 1 figure) A forc

Question

A block of mass 1.00 sits on an inclined plane as shown. (Intro 1 figure) A force of magnitude 50.0 is pulling the block up the incline. The coefficient of kinetic friction between the plane and the block is 0.500. The inclined plane makes an angle 10.0 with the horiz


1-What is the total work done on the block by the force of friction as the block moves a distance 8.00 up the incline?

2-What is the total work done on the block by the applied force as the block moves a distance 8.00 up the incline?

3-What is the total work done on the block by the force of friction as the block moves a distance 8.00 down the incline?

4-What is the total work done on the box by the applied force 50.0 in this case?

Explanation / Answer

1- Your data for the distance is garbled for some reason. I'll assume you intend it to say 8 m.
The friction is given by
Ff = mu*N =mu*m*g*cos10=0.500*8*9.8*cos10=38.60
where mu = 0.500 m = 1 kg, and g = 9.8 m/s^2.

The work done by the friction is
W_fric = Ff*d=38.60*8=308.87
where d = 8m

2- W_vec = F*d=50*8=400
where F = 50 N

3- Direction doesn't matter. The friction does work given by the magnitude of its force and the distance. The magnitude of its force doesn't change.

4- This also doesn't change

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.