A 280 kg piano slides 43 m down a 30 degree incline and is kept from accelerating by a man who is pushing back on it parallel to the incline Fig 6-36?

Answer 1

a.) the force exerted by the man is found by doing:

Fnormal= mass(gravity)(sin30) - u(mass)(gravity)(cos30)

(The "u" is the coefficient of friction)

b.) the work done by the man is found by using the force we found in the question above and putting it into:

W=Fd

My coeffiecent of friction was 0.40 so my total force was 421.5N

I just put it into W=421.5N(whateverdistanceIgiven)

for example is it was 9cm: (9cm = 0.09m)

W=421.5(0.09m) = 37.935J

c.) the work done by the friction force is solved the exact same way:

Since my coefficient of friction = 0.40 my Force of Friction = 0.41(280kg)(9.8m/s2)(cos30) = 950.5N

Then:

W=Fd = 950.5N(0.09m) = 85.545J

d.) The work done by gravity: First, find force of gravity by: mass(gravity)(sin30) = 1372N, then just plug into W=Fd = 1372(0.09m) = 123.48J

e.) the net work done on the piano = zero

See: 123.48J - 85.545J = 37.935J !

Answer 2

Draw your force diagram, label the 3 factors and use your formulas

F = m*a