Question

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 coefficient 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?

Answer 1

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:

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

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

$W_{total} = 131.314\,J$