A 30.0-kg crate is initially moving with a velocity that has magnitude 3.90 m/s in a direction 37.0 west of north. How much work must be don

Question

A 30.0-kg crate is initially moving with a velocity that has magnitude 3.90 m/s in a direction 37.0 west of north. How much work must be done on the crate to change its velocity to 5.62 m/s in a direction 63.0° south of east?

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Thu Thủy 3 years 2021-08-16T01:19:48+00:00 1 Answers 95 views 0

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  1. Orla
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    2021-08-16T01:21:04+00:00
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    Answer:

    Explanation:

    mass of crate, m = 30 kg

    initial velocity, u = 3.9 m/s at an angle 37° west of north

    final velocity, v = 5.62 m/s at an angle 63° south of east

    By teh work energy theorem, the work done is equal to the change in kinetic energy of the body.

    W = \frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

    W = 0.5 x 30 x (5.62² – 3.9²)

    W = 15 x 16.37

    W = 245.6 J

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