Question

A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of 16.0 m. She pushes in a direction 21.0° below the horizontal. A 46.0-N frictional force opposes the motion of the cart.

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

(b) Determine the work done by the pushing force.
____ J

(c) Determine the work done by the frictional force.
____ J

(d) Determine the work done by the gravitational force.
_____ J

Answer 1

Given

Mass = 16 kg

distance = 16 m

angle = 21^o

Frictional force   f = 46 N

(a)

apply Newton's second law of motion ,     

net force F = F*cos21 - f = 0

F = f/cos21

F = 46/cos21

F = 49.30 N

(b)

Work done by the pushing force

Work done W1 = F*d *cos0

W1 = (46)*(16)(cos0)

W1 = 736 J

(c)

work done by frictional force

W = f*d*Cos180

W = (46)*(16)(-1)

W  = - 736 N

(d)

Work done by Gravity

W_g = F*d*cos90

W_g = (mg)*(0)

W_g = 0 J