Question

A body of mass 2kg is projected vertically up with a velocity 5m/s.the work done on the body by gravitational force before it is brought to rest momentarily is

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A body of mass 2kg is projected vertically up with a velocity 5m/s. the work done on the body by gravitational force before it is brought to rest momentarily is?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• $\large{\text{Mass of The Body (m) = 2Kg.}}$
• $\large{\text{Initial velocity (u) = 5 m/s. }}$
• $\large{\text{Final velocity (v) = 0 m/s. }}$
• $\large{\text{Acceleration due to gravity (a) = - g.}}$

$\huge{\bold{\underline{Explanation:-}}}$

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$\large{\text{Applying Third kinematic equation}}$

$\large{\boxed{\tt v^2 - u^2 = 2as}}$

$\large{\text{Substituting the Values}}$

$\large{\tt \leadsto 0 - u^2 = 2 \times- g \times H_{max}}$

$\large{\tt \leadsto - u^2 = - 2gH_{max}}$

$\large{\tt \leadsto \cancel{-} u^2 = \cancel{-} 2gH_{max}}$

$\large{\tt \leadsto u^2 = 2gH_{max}}$

$\large{\leadsto {\underline{\underline{\tt H_{max} = \dfrac{u^2}{2g}}}}}$

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$\large{\underline{\underline{\text{Applying work Done Formula}}}}$

$\large{\boxed{\tt W = mgh}}$

$\large{\text{ Substituting the values}}$

$\large{\tt \leadsto W = mgH_{max}}$

$\large{\tt \leadsto W = mg \times \dfrac{u^2}{2g}}$

$\large{\tt \leadsto W = 2 \times 10 \times \dfrac{(5)^2}{2 \times 10}}$

$\large{\tt \leadsto W = 2 \times \cancel{10} \times \dfrac{(5)^2}{2 \times \cancel{10}}}$

$\large{\tt \leadsto W = 2 \times \dfrac{25}{2}}$

$\large{\tt \leadsto W = \cancel{2} \times \dfrac{25}{\cancel{2}}}$

$\huge{\boxed{\boxed{\tt W = 25 \: J}}}$

$\large{\underline{\underline{\text{So, the Work done is 25 Joules.}}}}$

Note:-

• Here Acceleration due to gravity is taken as negative as the Ball is thrown upward which is against the Gravitational force (or) Pull.

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Answer 2

Answer:

Given:

Mass of object = 2 kg

Initial velocity = 5 m/s

Final velocity = 0 m/s (as the object comes to rest )

To find:

Work done by Gravitational force before the object comes to rest momentarily.

Concept:

Work done is force multiplied with displacement.

So using the equations of kinematics, we can find out the displacement of the object.

And we already know that the force is due to gravity

Calculation:

v² = u² - 2gh

=> 0² = 5² - 2 × 10 × h

=> 0 = 25 - 20h

=> 20h = 25

=> h = 25/20

=> h = (5/4)

Work done = F × h

=> W = (m × acc.) × h

=> W = (2 × 10) × (5/4)

=> W = 25 J.

So work done = 25 J.