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26 June, 15:30

2.

A sled with a boy on it has a combined mass of m = 75 kg. As they slide across a horizontal surface of an ice-

covered lake, they come to rest in a distance of s = 25 m. If the coefficient of friction between the sled and the ice

is u = 0.25, calculate the amount of work done on the sled by friction. Keep in mind - negative Work decreases

Kinetic Energy whereas Positive Work increases it. So, don't forget a negative (-) sign in front of your answer.

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Answers (1)
  1. 26 June, 17:39
    0
    The work done on the sled by friction, W = - 4593.75 J

    Explanation:

    Given data,

    The combined mass of sled and the boy, m = 75 kg

    The displacement of the boy, S = 25 m

    The coefficient of the friction, u = 0.25

    The frictional force acting on the boy,

    F = u η

    Where,

    η - is the normal force acting on the boy (mg)

    Substituting the values,

    F = 0.25 x 75 x 9.8

    = 183.75 N

    Since the direction of the frictional force is against the direction of motion

    F = - 183.75 N

    The work done on the sled by friction,

    W = F x S

    = - 183.75 x 25

    = - 4593.75 J

    Hence, the work done on the sled by friction, W = - 4593.75 J
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