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15 April, 01:49

across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 165 N directed 26 degrees below the horizontal. While you slide the chair a distance of 6.00 m, the chair's speed changes from 1.30 m/s to 2.50 m/s. Find the work done by friction on the chair.

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Answers (2)
  1. 15 April, 02:18
    0
    W = - 847J

    Explanation:

    Given m = 18.8kg, F = 165N, θ = - 26° (below the horizontal, s = 6.0m, u = 1.30m/s and v = 2.50m/s

    In this problem, two forces act on the chair; the forward force F and the frictional force f. We would apply newton's second law to find the frictional force f after which we can calculate the workdone by the frictional force f*s.

    But for us to apply newton's second law, we need to know the acceleration of the chair cause by the net force.

    From constant acceleration motion equations

    v² = u² + 2as

    2.5² = 1.30² + 2a*6

    6.25 = 1.69 + 12a

    12a = 6.25 - 1.69

    12a = 4.56a

    a = 4.56/12

    a = 0.38m/s

    By newton's second law the net sum of forces equals m*a

    The force F has horizontal and vertical and components. It is the horizontal component of this force that pushes the chair against friction.

    Fx and f are oppositely directed.

    So

    Fx - f = ma

    165cos (-26) - f = 18.8*0.38

    148.3 - f = 7.14

    f = 148.3 - 7.14

    f = 141.2N

    Workdone = - fs = - 141.2*6.00 = - 847J

    W = - 847J

    Work is negative because it is done by a force acting on the chair in a direction opposite (antiparallel) to that of the intended motion.
  2. 15 April, 04:34
    0
    -847.2J

    Explanation:

    First find the acceleration from v^2 = u^2 + 2as

    v = 2.5 m/s

    u = 1.3 m/s

    a?

    s=6.00

    a = v^2-u^2/2s

    a = (2.5) ^2 - (1.3) ^2/2 * 6

    a = 0.38ms^-2

    From Newtons second law:

    (Force applied cos Θ) - (Frictional force) = ma

    Frictional force = ma - (Force applied cos Θ)

    Frictional force = (18.8*0.38) - (165 cos 26°)

    Frictional force = 7.144 - 148.3 = - 141.2N

    Therefore,

    Work done by friction = Frictional force * distance covered

    = - 141.2N * 6 = - 847.2J
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