Question

A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.50 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do

Answer 1

Answer:

1098.65 J

Explanation:

he desk moves at constant speed, this means that its acceleration is zero, so for Newton's second law the resultant of the forces acting on the desk is zero.

There are only two forces acting on the desk: the force of the student pushing it (F) and the frictional force Ff, acting in the opposite direction, so Netwon's second law becomes

$F-F_f=0\\so\\F=F_f$

the frictional force is

$F_f = \mu_k mg$

where $\mu_k = 0.400$  is the coefficient of kinetic friction, m=80.0 kg is the desk mass and g is the gravitational acceleration. From this equation, we find the intensity of the force:

$F= \mu_k mg\\\\=(0.400 )(80.0 kg)(9.81 m/s^2)\\\\=313.9 N$

finally, the work done by the student is the force times the distance across which the desk has been moved:

$W=Fd\\\\=(313.9 N)(3.50 m)\\\\=1098.65J$