Daniel B. answered • 10/30/20
A retired computer professional to teach math, physics
Let
Fa = 150N be the applied force
Fr be the force of friction (to be calculated later)
Wa be the work done by the applied force
Wr be the work done by the force of friction
m = 40.5 kg be a mass of the box
s = 6m be the length of the trajectory
f = 0.3 be the coefficient of friction
g = 9.8 m/s2 be the gravitational acceleration
v be the final speed of the box (to be calculated)
(a) In general, work done by a force is the dot product between
the force and its trajectory.
Since the force Fa and the trajectory s are parallel
we can simply multiply their magnitudes:
Wa = Fa . s = 150N . 6m = 900 J
(b) Energy due to friction is equal the work done by
the force of friction Fr.
That force is the product of the box's weight and the coefficient of friction
Fr = m . g . f
Wr = Fr . s = m . g . f . s = 40.5 kg . 9.8 m/s2 . 0.3 . 6m = 714.42 J
(c) Since the normal force is perpendicular to the trajectory s,
their dot product is 0.
So the work done by the normal force is 0.
(d) For the same reason the work done by the gravitational force is 0.
(e) The final kinetic energy is derived from the work of the applied force
minus the energy lost due to friction.
So the change in kinetic energy is Wa - Wr = 900 J - 714 J = 186 J
(f) Since the box is initially at rest, the change in kinetic energy
is equal the total final kinetic energy, which is
m . v2/2, and we know the value of that from (e)
m . v2/2 = Wa - Wr
v = sqrt(2 . (Wa- Wr)/m) = sqrt(2 . 186 / 40.5) = 3.02 m/s
