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blsea [12.9K]
3 years ago
9

A 12.0-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0∘ below the horizontal. The coeffici

ent of kinetic friction between the package and the chute’s surface is 0.40. Calculate the work done on the package by ___________.
(a) friction,
(b) gravity, and
(c) the normal force.
(d) What is the net work done on the package?
Physics
1 answer:
Fantom [35]3 years ago
6 0

Answer:

a) W_{f} = -56.659\,W, b) W_{g} = 187.973\,J, c) W_{N} = 0\,J, d) W_{total} = 131.314\,J

Explanation:

a) The friction force is:

f = \mu_{k}\cdot m\cdot g\cdot \cos \theta

The work done by friction is:

W_{f}=-\mu_{k}\cdot m \cdot g \cdot \cos \theta\cdot \Delta s

W_{f} =- (0.4)\cdot (12\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\cos 53^{\textdegree})\cdot (2\,m)

W_{f} = -56.659\,W

b) The work done by gravity is:

W_{g} = m\cdot g \cdot \sin \theta \cdot \Delta s

W_{g} = (12\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\sin 53^{\textdegree})\cdot (2\,m)

W_{g} = 187.973\,J

c) As normal force has a direction perpendicular to motion direction, the work done is equal to zero:

W_{N} = 0\,J

d) The net work done on the package is:

W_{total} = W_{f} + W_{g} + W_{N}

W_{total} = -56.659\,J +187.973\,J + 0 J.

W_{total} = 131.314\,J

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