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.
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.
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.
Repeat all calculations.
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