Question

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))

Answer 1

Answer:

400 Nm

Hope this answer is helpful.

Answer 2

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.

#SPJ3