Question

a body weighing 1000kg accelerates uniformly from 10 m/s to 20 m/s .Find the amount of work done during this period. hint - {work done = change in kinetic energy}​

Answer 1

Explanation:

1/2 mv^2 - 1/2 mu^2 is the change in K.E

mass = 1000kg

v= 10m/s

u= 20m/s

so

$\frac{1}{2} \times 1000 \times {20}^{2} - \frac{1}{2} \times 1000 \times {10}^{2}$

$\frac{1}{2} \times 1000( {20}^{2} - {10}^{2} )$

$500 (400 - 100) \\ 500 \times 300 = 150000 \\ 1.5 \times {10}^{5}$

J

Answer 2

Answer:1,50,000 J

Explanation:

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