A person pushes a 11.3-kg shopping cart at a constant velocity for a distance of 19.4 m on a flat horizontal surface. She pushes in a direction 29.2 ° below the horizontal. A 44.6-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts?

Answer 1

Given: m=11.3kg. A = 29.2 deg., Ff = 44.6N.

Wc = mg = 11.3kg * 9.8N/kg = 110.74N. =
Weight of cart.
Fc = (110.74N,0deg.).
Fp = 110.74sin(0) = 0 = Force parallel to surface.
Fv = 110.74cos(0) = 110.74N. = Force
perpendicular to surface.

Ff = 44.6N = Force of fiction.

Fn=Fap*cos29.2 - Fp - Ff = ma = 0, a=0.
Fap*cos29.2 - 0 - 44.6 = 0,
Fap*cos29.2 = 44.6,
Fap = 44.6 / cos29.2 = 51.1N. = Force
applied.