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14 April, 00:48

A 19.0-kg cart is moving with a velocity of 7.20 m/s down a level hallway. A constant force of - 13.0 N acts on the cart and its velocity becomes 3.60 m/s.

a) What is the change in kinetic energy of the cart? - 369.36

b) How much work was done on the cart?

c) How far did the cart move while the force acted? m

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Answers (2)
  1. 14 April, 02:58
    0
    a. - 369.36J

    b. - 123.9J

    c. 9.52m

    Explanation:

    From the expression for kinetic energy

    K. E=1/2mv^2

    Since the mass is constant, but the velocity changes. Hence the change in kinetic energy is

    K. E=1/2*19 (3.6²-7.2²)

    K. E = - 369.36J

    b. to determine the workdone by the force, we determine the distance moved.

    But the acceleration is from

    F=ma,

    a=f/m

    a=-13/19

    0.68m/s²

    the distance moved is

    s=v²/2a

    s=3.6²/2*0.68

    s=9.52m

    Hence the work done is

    W=force * distance

    W=-13*9.52

    W=-123.9J

    d. the distance moved is

    s=v²/2a

    s=3.6²/2*0.68

    s=9.52m
  2. 14 April, 03:17
    0
    a. - 369.36J

    b. - 369.36J

    c. 28.41m

    Explanation:

    a. ΔK. E = 1/2 * m * (v₂²-v₁²)

    ΔK. E = change in kinetic energy

    v₂ = final velocity = 3.6m/s

    v₁ = initial velocity = 7.2m/s

    m=mass=19kg

    ΔK. E = 1/2 * 19 * (3.6²-7.2²)

    =1/2*19 * (12.96-51.84)

    =9.5 * (-38.88)

    = - 369.36J

    b. the workdone is equal to the change in kinetic energy

    ∴ W. D = - 369.36J

    c. W. D = force * distance

    distance = W. D:force

    = - 369.36 : - 13.0

    =28.41m
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