Question

A force of 200 newton applied standing body with moves 10m/s and stopped 5second .calculate the work done.​

Answer 1

Answer:

Given

• force = 200 N
• initial velocity= 10 m/sec
• final velocity= 0
• time taken = 5 sec

To find

work done

Solution

Work done of an object is defined as magnitude of the force multiplied by the distance moved the object in direction of which force is applied.

${ \underline{ \boxed{ \sf{w = f.s \: cos\theta}}}}$

where

• f denotes force applied
• s denotes the distance
• w denotes the work done

In order to find the work done we need to first calculate the distance .

In order to find the work done we need to first calculate the distance .To find the distance we need to find the acceleration though the 1st equation of motion.

1st equation of motion is given by -:

${ \underline{ \boxed{ \sf{v = u + at}}}}$

where

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time.

Substitute the values

➝10 + a× 5

➝-10 = 5a

➝a= -10 /5

➝a= -2 m/sec ²

according to 3rd equation of motion

${ \underline{ \boxed{ \sf{ {v}^{2} = {u}^{2} + 2as}}}}$

➝0= (10)²+ 2× (-2) × s

➝0 = 100 -4s

➝-100 = -4s

➝s= 100/4

➝s= 25 meter

substitute the the distance in the formula of work

$\sf{➝w = 200 \times 25} \\ \sf{ ➝w = 5000 \: joule} \\ \sf{➝w = 5 \: kilojoules}$

Answer 2

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Given:-

force = 200 N

initial velocity= 10 m/sec

final velocity= 0

time taken = 5 sec

To find:-

work done = ?

Solution :-

~Work done is defined as product of the force and the distance over which the force is applied.

${ \underline{ \boxed{ \sf{w = f.s \: cos\theta}}}} </p><p>w=f.scosθ$

where

f denotes force applied

s denotes the distance

w denotes the work done

1st equation of motion is given by -:

${ \underline{ \boxed{ \sf{v = u + at}}}}$

where

v denotes final velocity

u denotes initial velocity

a denotes acceleration

t denotes time.

Put the values

➝1Woh 0 + a× 5

➝-10 = 5a

➝a= -10 /5

➝a= -2 m/sec ²

According to 3rd equation of motion

${ \underline{ \boxed{ \sf{ {v}^{2} = {u}^{2} + 2as}}}}$

v 2 =u2 +2as

0= (10)²+ 2× (-2) × s

0 = 100 -4s

-100 = -4s

s= 100/4

s= 25 meter

substitute the the distance in the formula of work

w = 200times × 25 = 5000 joule = 5 kilojoules

W = 5 kilojoules