Physics, asked by mitparikh6423, 9 months ago

A body of mass 2 kg is dropped from rest from a height 20 m from the surface of Earth. The body hiys the ground with velocity 10 m/s, then work done (g=10m//s^(2))

Answers

Answered by rupali8153gmailcom2
2

Answer:

400 Nm

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Answered by archanajhaasl
0

Answer:

The work done is 500J.

Explanation:

The total work done by the body is given as,

W=P.E.+K.E.         (1)

Where,

W=total work done by the body

P.E.=Potential energy

K.E.=Kinetic energy

From the question we have,

The mass of the body(m)=2 kg

The height from which the body is dropped(h)=20m

The velocity with which the body hits the ground(v)=10m/s

The acceleration due to gravity(g)=10m/s²

The kinetic energy(K.E.)

K.E.=\frac{1}{2}mv^2        (2)

By inserting variables in equation (2) we get;

K.E.=\frac{1}{2}\times 2\times 10^2

K.E.=100J   (3)

The potential energy(P.E.)

P.E.=mgh      (4)

By inserting variables in equation (4) we get;

P.E.=2\times 10\times 20

P.E.=400J   (5)

By substituting the values of "K.E." and "P.E." in equation (1) we get;

W=100+400

W=500J  (6)

Hence, the work done is 500J.

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