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AlexFokin [52]
4 years ago
13

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?
Physics
1 answer:
Elodia [21]4 years ago
7 0

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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