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2 January, 01:20

A 2.41 kg block is pushed 1.42 m up a vertical wall with constant speed by a constant force of magnitude f applied at an angle of 59.9 ◦ with the horizontal. the acceleration of gravity is 9.8 m/s 2. 2.41 kg 59 f. 9 ◦ if the coefficient of kinetic friction between the block and wall is 0.521, find the work done by f.

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  1. 2 January, 04:25
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    Refer to the diagram shown below.

    The weight of the block is

    W = (2.41 kg) * (9.8 m/s²) = 23.618 N

    The kinetic coefficient of friction is

    μ = 0.521

    The normal reaction is

    N = f cos (59.9°) = 0.5015f N

    The frictional resistive force is

    R = μN = 0.521 * (0.5015f N) = 0.2613f N

    For dynamic force balance,

    f sin (59.9°) = W + R

    0.8652f = 23.618 + 0.2613f

    0.6039f = 23.618

    f = 39.109 N

    The block moves by 1.42 m.

    The work done is

    W = (f sin (59.9° N) * (1.42 m)

    = 39.109*sin (59.9°) * 1.42

    = 48.05 J

    Answer: 48.0 J (nearest tenth
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