Asked by Josh

A 275 kg piano slides 4.0 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.
a) Calculate the magnitude of the force exerted by the man.

b) Calculate the work done by the man on the piano.

(c) Calculate the work done by the friction force.

Answers

Answered by Henry
a. Wt.=m*g = 275kg * 9.8N/kg=2695 N =
Wt of piano.

Fp = 2695*sin30 = 1348 N. = Force parallel to incline.
Fn = 2695*cos30 = 2334 N. = Normal = Force perpendicular to incline

Fk = u*Fn = 0.40 * 2334 = 933.6 N. =
Force of kinetic friction.

Fex-Fp-Fk = m*a
Fex-1348-933.6 = m*0 = 0
Fex - 2282 = 0
Fex = 2282 N. Exerted by man.

c. Work = Fk*d = 933.6 * 4 = 3734 J.

Answered by henry2,
Correction: Fex-Fp+Fk = M*a.
Repeat all calculations.
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