Question

In: Physics

A person pushes a 14.6-kg shopping cart at a constant velocity for a distance of 26.8...

A person pushes a 14.6-kg shopping cart at a constant velocity for a distance of 26.8 m on a flat horizontal surface. She pushes in a direction 26.8 ° below the horizontal. A 39.0-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.

Solutions

Expert Solution

(a)

Apply Newton's second law of motion, Net force is,

               ΣF = F cos26.8 - f = 0

               F cos26.8 - f = 0

                             F   = f / cos26.8

                     Fshopper= 43.69 N

(b)

Work done by the pushing force is,

W1 = Fd cos26.8

                  = (39)(26.8)( cos26.8 )

                   ≈ 932 J

(c)

Work done by frictional force is,

W = fd Cos180

                = (39)(26.8)(-1)

                = - 1045.2 J

(d)

Work done by Gravity is,

Wg = Fd cos90

                   = (mg)(0)

                   = 0 J

genius_generous answered 4 weeks ago
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