Question

A particle of mass 100 g is thrown vertically upwards with a speed of 5 m//s. The work done by the force of gravity during the time the particle goes up is

Answer 1

Answer:

Given:

Mass of object = 100 g

Initial Velocity = 5 m/s

To find:

Work done by force of gravity during the time particle goes up.

Concept:

As per Work-Energy Theorem , we can say that ; Work done by all forces is equal to the change in Kinetic energy during the motion.

$\boxed{\huge{\red{\sf{ Work = \Delta KE}}}}$

Calculation:

Initial Kinetic Energy = ½ m (v1)²

=> KE1 = ½ × 0.1 × (5)²

=> KE1 = 1.25 J

Final Kinetic Energy = ½ m (v2)²

=> KE2 = ½ × 0.1 × (0)²

=> KE2 = 0 J

So work done :

Work = KE2 - KE1

=> Work = 0 - 1.25

=> Work = - 1.25 J

So final answer is :

$\boxed{\sf{ \green{ \huge{work = -1.25 \: Joule}}}}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

$\sf Given \begin{cases} \rm{Mass \: of \: Object \: (m) \: = \: 100 \: g \: = \: 0.1 \: kg} \\ \rm{Final \: velocity \: (v) \: = \: 0 \: ms^{-1}} \\ \rm{Initial \: velocity \: (u) \: = \: 5 \: ms^{-1}} \\ \rm{Accleration \: (g) \: = \: - \: 10 \: ms^{-2}} \\ \rm{Work \: Done \: = \: ?} \end{cases}$

Use Third equation of Motion :

$\large{\boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}} \\ \\ \implies {\sf{0^2 \: - \: 5^2 \: = \: \times \: 2 \: \times \: -10 \: \times \: s}} \\ \\ \implies {\sf{-25 \: = \: -20s}} \\ \\ \implies {\sf{s \: = \: \dfrac{25}{20}}} \\ \\ \implies {\sf{s \: = \: 1.25}} \\ \\ {\underline{\sf{\therefore \: Height \: till \: particle \: goes \: is \: 1.25 \: m}}}$

_______________________________

We know that :

$\large{\boxed{\sf{W \: = \: mgh}}} \\ \\ \implies {\sf{W \: = \: 0.1 \: \times \: -10 \: \times \: 1.25}} \\ \\ \implies {\sf{W \: = \: -1 \: \times \: 1.25}} \\ \\ \implies {\sf{W \: = \: -1.25}} \\ \\ {\underline{\sf{\therefore \: Work \: Done \: is \: -1.25 \: J}}}$