Question

A tension force of 165 N inclined at 15.0° above the horizontal is used to pull a 33.0 kg shipping crate a distance of 3.60 m on a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b) the coefficient of kinetic friction between the crate and surface.

HINT

(a)

the work done by the tension force (in J)

J

(b)

the coefficient of kinetic friction between the crate and surface

Answer 1

Part A.

Work-done by force 'F' over displacement 'd' is given by:

W = F.d = F*d*cos

F = Tension force = 165 N

d = displacement = 3.60 m

= Angle between tension force and displacement = 15.0 deg

So,

W = 165*3.60*cos 15.0 deg

W = 573.76 J = 574 J = Work done by tension force

Part B.

Friction force on crate will be given by:

Ff = *N

N = Normal force

from FBD:

N = W - F*sin = m*g - F*sin

So,

Ff = *(m*g - F*sin )

Now given that crate is moving at constant speed, So net force on crate is zero, that means

In horizontal direction:

F_net = F*cos - Ff = 0

Ff = F*cos

So,

F*cos = *(m*g - F*sin )

= F*cos /(m*g - F*sin )

So,

= 165*cos 15 deg/(33.0*9.81 - 165*sin 15 deg)

= Coefficient of kinetic friction = 0.567

Let me know if you've any query.