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13 January, 14:22

You observe someone pulling a block of mass 41 kg across a low-friction surface. While they pull a distance of 5 m in the direction of motion, the speed of the block changes from 5 m/s to 6 m/s. Calculate the magnitude of the force exerted by the person on the block.

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Answers (2)
  1. 13 January, 14:38
    0
    45.1 N

    Explanation:

    Using Newton's Fundamental equation of Kinematics,

    F = ma ... Equation 1

    Where F = force exerted by the person on the block, m = mass of the block, a = acceleration of the block

    We can look for the acceleration of the block by applying newton's equation of motion,

    Using,

    v² = u²+2as ... Equation 2

    Where v = final velocity, u = initial velocity, a = acceleration, s = distance.

    make a the subject of the equation

    a = (v²-u²) / 2s ... Equation 3

    Given: v = 6 m/s, u = 5 m/s, s = 5 m

    Substitute into equation 3

    a = (6²-5²) / (2*5)

    a = 11/10

    a = 1.1 m/s²

    Also Given: m = 41 kg

    Substitute into equation 1

    F = 41 (1.1)

    F = 45.1 N
  2. 13 January, 14:41
    0
    Answer: 45.1N

    Explanation:

    Answer: 45.1N

    Explanation:

    The work done in the block is equal to change in energy is the system. We then go ahead to calculate W with the given values of initial and final speed.

    W = Ef - Ei

    W = 1/2MVf² - 1/2MVi²

    W = 1/2M (Vf² - Vi²)

    W = 1/2 * 41 (6²-5²)

    W = 1/2 * 41 * 11

    W = 225.5J

    We note the definition of work and solve for F.

    Taking into cognizance the displacement we have been given in the question. We then have

    F = W/x

    F = 225.5/5

    F = 45.1N
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