physics Problem 6.8 A 320kg piano slides 3.6m down a 33 incline and is kept from
ID: 1311226 • Letter: P
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physics
Problem 6.8 A 320kg piano slides 3.6m down a 33 incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (see the figure(Figure 1) The effective coefficient of kinetic friction is 0.38. Figure 1 v of 1 Part A Calculate the force exerted by the man Express your answer using two significant figures reset shortcuts help Submit My Answers Give U Part B Calculate the work done by the man on the piano. Express your answer using two significant figures reset shortcuts help Submit My Answers Give UExplanation / Answer
1. Calculate the Friction Force
Ff = Cf cos theta m g
Ff= Cf cos33 m g
2. Calculate Force of Acceleration on Piano
Fg = sin33 mg
a. Force exerted by man:
You are in equilibrium -- constant velocity.
Fg = Ff + Fman
sin33 mg = Cf cos33 mg + Fman
Fman = mg (sin33 - Cf cos 33)
Fman = 320kg * 9.81 m/s^2 (0.5446 - (0.38 * 0.83867))
Fman = 709.28 N
(b) Work done by man:
W = Fman *d
d= - 3.6 m ........ note the direction is opostite the force
W = -2553.42 J
(c) Work done by friction force
W = Ff * d
W = Cf cos33 mg * -3.6
W = -3601.608 J
(d) Work done by gravity
W = Fg * d ... Here d is in the direction of acceleration
W = mg sin33 * 3.6
W = 6155.03 J
(e) W = 6155.03 - 3601.608 - 2553.42 = 0
No surprize. The piano maintains constant velocity.
6.18) W = change in KE
w = 0.5*1250*30.55^2
W = 583526.2J
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