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Calculating Forces & Work for 330-kg Piano on 28° Incline
Chapter 6, Problem 8P(choose chapter or problem)
A 330-kg piano slides 3.6 m down a \(28^{\circ}\) incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). The effective coefficient of kinetic friction is 0.40. Calculate:
(a) the force exerted by the man,
(b) the work done by the man on the piano,
(c) the work done by the friction force,
(d) the work done by the force of gravity, and
(e) the net work done on the piano.
Questions & Answers
QUESTION:
A 330-kg piano slides 3.6 m down a \(28^{\circ}\) incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). The effective coefficient of kinetic friction is 0.40. Calculate:
(a) the force exerted by the man,
(b) the work done by the man on the piano,
(c) the work done by the friction force,
(d) the work done by the force of gravity, and
(e) the net work done on the piano.
Step 1 of 5
The mass of the piano is: \(m=330 \mathrm{~kg}\).
It slides down for 3.6 meters.
The angle of inclination is: \(\theta=28^{\circ}\).
The effective coefficient of kinetic friction is \(\mu_{k}=0.40\).
a) The normal reaction is equal to,
\(N=m g \cos \theta\)
For the x direction, the force equation will be,
\(m g \sin \theta-F_{P}-F_{\text {fric }}=0\)
\(\begin{aligned} \Rightarrow F_{P} & =m g \sin 28^{\circ}-\mu_{k} N \\ & =m g \sin 28^{\circ}-\mu_{k} m g \cos \theta \\ & =330 \times 9.8(0.4694-0.40 \times 0.8829) \\ & =375.92 \mathrm{~N} . \end{aligned}\)
So, the force exerted by the person is \(375.92 \mathrm{~N}\).
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Calculating Forces & Work for 330-kg Piano on 28° Incline
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In this video we break down a complex physics problem about a 330-kg piano sliding down a 28° incline while a man resists its motion. We meticulously calculate the force exerted by the man work done by various forces including friction and gravity and the net work done on the piano.