A person pushes a 24.2-kg shopping cart at a constant velocity for a distance of 17.6...

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A person pushes a 24.2-kg shopping cart at a constant velocityfor a distance of 17.6 m on a flat horizontal surface. She pushesin a direction 22.7 Â° below the horizontal. A 42.4-N frictionalforce opposes the motion of the cart. (a) What isthe magnitude of the force that the shopper exerts? Determine thework done by (b) the pushing force,(c) the frictional force, and (d)the gravitational force.

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4.0 Ratings (580 Votes)

(a)

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Î£F = F cos17.6 - f = 0

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â F cos17.6 - f = 0

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â FÂ Â = f / cos17.6

=42.4/0.9532

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â F_shopper= 44.48 N

(b)

Work done by the pushing force is,

W1 = Fd cos17.6

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (42.4)(17.6)( cos17.6 )

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â‰ˆ 711 J

(c)

Work done by frictional force is,

W = fd Cos180

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (42.4)(17.6)(-1)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = - 746.24 J

(d)

Work done by Gravity is,

W_g = Fd cos90

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (mg)(0)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0 J

I hope help you.....!!

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A person pushes a 24.2-kg shopping cart at a constant velocity for a distance of 17.6...

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