A person pushes a 15.7-kg shopping cart at a constant velocity for a distance of 25.9 m on a flat horizontal surface. She pushes in a direction 23.7 ° below the horizontal. A 32.7-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

a)    F = 35.7 N, b)   W = 846.7 J, c)   W = - 846.9 J, d) W=0

Explanation:

a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally

let's break down the pushing force.

        cos (-23,7) = Fₓ / F

        sin (-237) = F_y / F

        Fₓ = F cos 23.7 = F 0.916

        F_y = F sin (-23.7) = - F 0.402

Y axis  

       N- W - F_y = 0

       N = W + F 0.402

X axis

       Fₓ - fr = 0

       F 0.916 = fr

       F = fr / 0.916

       F = 32.7 / 0.916

       F = 35.7 N

It is asked to calculate several jobs

b) the work of the pushing force

       W = fx x

       W = 35.7 cos 23.7 25.9

       W = 846.7 J

c) friction force work

        W = F x cos tea

friction force opposes movement

        W = - fr x

         W = - 32.7 25.9

         W = - 846.9 J

d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero

          W = 0

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures