Question

Sample Test Paper, 2011
cience
At a height of 9m from ground if a body of mass 2 kg has gravitational potential energy 36J, what
will be its final velocity with which it strikes the ground if it is dropped from there?
B. 6 V5 m/sec​

Answer 1

Answer:

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Explanation:

At a height of 9 m from ground if a body of mass 2 kg has gravitational potential energy 36J, what will be its final velocity with which it strikes the ground if it is dropped from there? B. 6 V5 m/sec​

Answer 2

Answer:

The velocity with which it strikes the ground if it is dropped from a height of 9m is 6m/s.

Explanation:

We will solve this question from energy conservation law.

• Which states that "Energy is capable of changing its forms".
• Here as per the question, the gravitational potential energy will be converted into potential energy as no external force is acting on the given system.

From the question, we have the following data,

The height from which the body is dropped(h)=9m

The mass of the body(m)=2kg

The gravitational potential energy(P.E)=36 J

And we know the kinetic energy is calculated as,

$\mathrm{K.E.=\frac{1}{2}mv^2 }$             (1)

Where,

K.E.=kinetic energy

m=mass of the body

v=velocity with which the body is moving

And after applying the energy conservation principle we have,

$\mathrm{K.E.=P.E}$         (2)

By inserting equation (1) and all the required values in equation (2) we get;

$\mathrm{\frac{1}{2}mv^2 =P.E}$

$\mathrm{\frac{1}{2}\times 2\times v^2 =36}$

$\mathrm{v^2=36}$

$\mathrm{v=\sqrt{36} }$

$\mathrm{v=6\ m/s }$

So, the velocity with which it strikes the ground if it is dropped from a height of 9m is 6m/s.

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