A box of mass 10.9 kg with an initial velocity

Question

of 2.01 m/s slides down a plane, inclined
at 38◦ with respect to the horizontal. The coefficient
of kinetic friction is 1. The box stops
after sliding a distance x. The acceleration
due to gravity is 9.8 m/s.

a. How far does the box slide?. The positive
x-direction is down the plane.
Answer in units of m

b.What is the the work done by friction?
Answer in units of J

c.What is the work done by the normal force?
Answer in units of J

d.What is the magnitude of the work done by gravity?
Answer in units of J

e.What is the magnitude of the instantaneous
power generated by friction half way between
the initial and final positions?
Answer in units of W

f.What is the magnitude of the average power
generated by friction from start to stop?
Answer in units of W

Answer 1

M*g = 10.9kg * 9.8N./kg = 106.82 N. = Wt
of the box.

Fp = 106.82*sin38 = 65.8 N. = Force parallel to the incline.
Fn = 106.82*Cos38 = 84.2 N. = Normal
force.

Fk = u*Fn = 1 * 84.2 = 84.2 N. = Force of kinetic friction.

Note: If the coefficient of kinetic friction is 1, the box cannot slide down the incline; because the kinetic
friction is higher than the force
parallel to the incline(Fp).