Question

a body 1000 kg accelerates uniformly from 10m/s to 20m/s find the amount of work done during this period

Answer 1

Given:
m= 1000kg
v1= 10 m/s
v2= 20m/s

Kinetic Energy of first velocity= 1/2×1000×(10)^2
=500×100
=50000J
K.E of second velocity=1/2×1000×(20)^2
=500×400
=200000J
Work done = change in kinetic energies
= 200000 - 50000
= 150000 Joules

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Answer 2

The amount of work done during this period is $W=15\times 10^4\ J$.

Explanation:

Given that,

Mass of the body, m = 1000 kg

Initial speed of the body, u = 10 m/s

Final speed of the body, v = 20 m/s

To find,

Amount of work done.

Solution,

According to work energy theorem, the work done by an object is equal to the change in its kinetic energy. It is given by :

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 1000 ((20)^2-(10)^2)$

W = 150000 J

$W=15\times 10^4\ J$

So, the amount of work done during this period is $W=15\times 10^4\ J$. Hence, this is the required solution.

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Work energy theorem

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