A 275 kg piano slides 4.2 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40

Question

(a) Calculate the force exerted by the man.

N
(b) Calculate the work done by the man on the piano.
J
(c) Calculate the work done by the friction force.
J
(d) What is the work done by the force of gravity?
J
(e) What is the net work done on the piano?
J

Answer 1

component of weight down slope = 275 * 9.81 * sin 30 = 1349 N
component of weight normal to slope = 275*9.81*cos 30 = 2336 N
so friction force up = .4 *2336 = 935 N
total force up = F + 935
total force down = 1349
no acceleration so force up = force down slope
F + 935 = 1349
solve for F in Newtons
now work = force * distance
F is up, motion is down so work done by man is -4.2 F Newton meters or Joules
The friction force, same deal
However gravity is component down slope times motion down slope so positive 1349 * 4.2

well gravity work better be equal and opposite to sum of man and friction :)

Answer 2

a. M*g = 275*9.8 = 2695 N. = Wt. of piano.
Fp = 2695*sin30 = 1348 N. = Force parallel with plane.
Fn = 2695*Cos30 = 2334 N. = Normal force.
u*Fn = 0.4*2334 = 934 N. = Force of kinetic friction.

Fe-Fp+u*Fn = M*a.
Fe-1348+934 = M*0,
Fe - 414 = 0,
Fe = 414 N. = Force exerted.

b. W = Fe*d = 414 * 4.2 = 1739 Joules.

c. W = uFn * d = 934 * 4.2 =

d. W = 1348 * 4.2 =

Answer 3

e. Fnet = Fe-Fp+uFn = 414-1348+934 = 0.
Wnet = Fnet * d = 0 * 4.2 = 0.