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? (Hint: Start by setting up a coordinate system and writing the vectors for the displacement and for each force. Then apply the expression for work when a constant force is applied.)

Answer 1

Answer:

a. 56.62J

b. 187.84J

c. 0J

d. 131.22J

Explanation:

#Use the attached figure for illustration:

Given the mass of package as 12 kg, distance as 2m and angle of inclination as 53°:

a. Kinetic friction equals normal force n which equals magnitude to the weight of the package, since package is moving along an incline.

-Kinetic friction force has the constant magnitude:

$f_k=\mu _kn\ \ \ \ \ \ \ \ \ \ \ ....i\\\\n=mg\ cos 53\textdegree\\\\f_k=\mu_kmg \ cos 53\textdegree \ \ \ \ \ \ \ .....ii\\\\f_k=0.4\times12kg\times9.8m/s^2cos 53\textdegree\\\\f_k=28.31N\ \ \ \ \ \ \ \ \ \ \ \ \ ....iii\\\\W=F.s=Fs\ cos \phi\ \ \ \ \ \ \ \ \ \ \ ....iv$

#if F and s is in the same direction the $\phi=0$. Friction is in the opposite direction, therefore $\phi=180\textdegree$:

$W_f=f_scos \ 180\textdegree\\\\=28.31N\times 2m\ cos 180\textdegree\\\\=-56.62J$

Hence, the friction force is -56.62J

b. From the question we calculate the angle at which work is done as 90°-53°=37°.

-Also, taking g=9.8m/s/s

The work done by gravity on the incline is calculated as:

$F_g=mg\\\\W_g=F_gs \ cos \ \phi\\\\=mg \ s \ cos (90\textdegree -53\textdegree)\\\\=12kg\times9.8m/s^2\times cos \ 37\textdegree\\\\=187.84J$

Hence, work done by gravity is 187.84J

c.Work done by the normal force on the package(from the chute's surface)is calculated as:

-Angle of work is 90°, given that the force is normal(perpendicular to the distance covered):

$W_n=F_ns\ cos \theta\\\\=F_ns \ cos90\textdegree\\\\=0J$

Hence, work done by the normal force is 0J

b. To find the network done on the package, we sum all individual works on the package:

-From a, Wf=-56.62J and from b the work due to gravity is 187.84J, and from c , 0J is due to the normal force:

$W_{net}=W_f+W_g+W_n\\\\=-56.62J+187.84J+0J\\\\=131.22J$

Hence, the net work done on the package is 131.22J