Frictional forces

Question

A shopper at the Fries’ market place pushes a 13.0-kg shopping cart at a constant velocity for a distance of 58.7 m on a level surface. He pushes in a direction 24.5⁰ below the horizontal. A 53.8 N frictional force opposes the motion of the cart.

(a) What is the magnitude of the force that the shopper exerts on the cart?

(b) What is the work done by the pushing force?

(c) What is the work done by the frictional force?

(d) What is the work done by the gravitational force? 

Answer 1

a) Since your at a constant velocity, your forces are balanced. This means that for the horizontal direction, your frictional force (F_f) is equal to the horizontal component of your applied force (F_a). So:

F_f = F_a cos(24.5)

Substituting the value for frictional force gives F_a a magnitude of 59.1 N

b/c) the equation for work is W = F*d*cos(θ) with the angle being between the force and displacement vectors. For the applied force, this is the 24.5° given and for friction is 180°. Therefore

W_A = 59.1 * 58.7 * cos(24.5) = 3160 N*m

W_f = 53.8 * 58.7 * cos(180) = -3160 N*m

d) Work is only done when the force is in the direction of motion. Since the force of gravity is perpendicular to the horizontal motion, it does zero work. Mathematically:

W_g = F_g * 58.7 * cos(90) = 0