8. A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.

Question

8. A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0º below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the shopper exerts, using energy considerations. (e) What is the total work done on the cart?

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Thạch Thảo 5 years 2021-09-01T16:06:39+00:00 1 Answers 8 views 0

Answers ( )

  1. thienthanh
    0
    2021-09-01T16:08:33+00:00
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    Answer:

    Part a) Work done by friction force

    W = -700 J

    Part b) Work done by gravity

    W = 0

    Part c) Work done by shopper

    W = 700 J

    Part d) Force exerted by shopper

    F = 38.6 N

    Part e) Total work done

    W = 0

    Explanation:

    Part a)

    Work done by friction force is given as

    W = -F_f (d)

    so we have

    W = -(35)(20)

    W = -700 J

    Part b)

    Since gravitational force is perpendicular to the displacement

    so we have

    W = Fd cos90

    W = 0

    Part c)

    Work done by the shopper is same as that of work done by friction force

    Because here trolley is moving with constant velocity

    So net force on it is zero

    W_{shopper} + W_{friction} = 0

    W_{shopper} = - W_{friction}

    W_{shopper} = 700 J

    Part d)

    F_f + Fcos25 = 0

    -35 + Fcos25 = 0

    F = 38.6 N

    Part e)

    Total work done is given as

    W = W_{shopper} + W_{friction}

    W = 0

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