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polet [3.4K]
3 years ago
10

A 3kg object has an initial velocity (6i - 2j) m/s (a ) what is its kinetic energy at this time? (b) Find total work done on the

object as its velocity changes to (8i+4j) m/s​?
Physics
1 answer:
guapka [62]3 years ago
3 0

Answer:

K.E =  \frac{1}{2} m {v}^{2}  \\  {v}^{2}_i  =  {v}^{2} _x + {v}^{2} _y \\  =  {(6)}^{2}  +  {( - 2)}^{2}  = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (40) = 60J \\  \\ {v}^{2}_f  =  {v}^{2} _x + {v}^{2} _y \\  =  {(8)}^{2}  +  {(4)}^{2}  = 80m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\  = 120J - 60J \\  = 60J

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<h3>Hooke's Law</h3>

Hooke's law states that the force applied to an elastic material is directly proportional to its extension, provided its elastic limit is not exceeded.

To calculate the amount of work done by Hooke's law, first, we need to find the force constant of the spring.

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Finally, To find the work done in stretching the spring from 50 cm to 80 cm, we use the formula below.

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Learn more about Hooke's law here: brainly.com/question/12253978

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