Question

A 2.98 X 102-kg crate is being pulled across a horizontal surface by a force F which makes an angle of 35.90above the horizontal. Suppose that the coefficient of kinetic friction is 0.222, what should the magnitude of F be so that the net work done by it and the kinetic frictional force is zero?

Answer 1

F = 690.127 N

Step-by-step explanation:

given,

mass of crate = 2.98 X 10² kg

angle with horizontal = 35.9°

kinetic friction = 0.222

Since there is no motion in the vertical direction , therefore

Rn + F Sin 35.9 = mg

Rn = (298 x 9.81) - F Sin 35.9

= 2923.38 - F Sin 35.9 ----------(1)

Friction force (f_k) = μ_k x Rn = 0.222 x (2923.38 - F Sin 35.9)

Now since the work done is ZERO therefore net force must be ZERO.

Hence

F Cos 35.9 - f_k = 0

F Cos 35.9 = f_k

f_k = 0.222 x (2923.38 - F Sin 35.9)

3.65 F = 2923.38 - 0.586 F

4.236 F = 2923.38

F = 690.127 N