Physics, asked by vgadagadi295, 2 months ago

A force of 10 N acts for 20 second on a body of mass 2 kg initially at rest. Calculate the energy required by the body and the work done by the applied force.​

Answers

Answered by Rajkd180203
2

"Answer:

1000J is the answer

Explanation:

F/m= a= 5m/s²

u=0 , t=20s

v= 5m/s² * 20 s

v= 100m/s

Change in KE = KEf-KEi

(KEi = 0 as it was initially at rest )

so change in KE= 1/2 * 2kg * 10000 m²/s²

so change in KE = 10000kgm²/s² = 10000J

but we Know Work done by all forces = change in KE = 1000J

"

Answered by anthonypaulvilly
1

Answer:

F = 10N

mass = 2kg

F = ma

a = F / m = 20 / 2 = 5m/s²

acceleration (a) = 5m/s²

time (t) = 20s

initial velocity (u) = 0

displacement (s) = ut + at²/2

s = 0 + 5×20² / 2

s= 5 × 20 × 20/2

s = 5 × 20 × 10  

s = 1000m

Work done = F × s

                   = 10×1000

                   = 10000j

v = u + at

v = 0 + 5×20

v = 100m/s

Kinetic energy = 1/2mv²

                        = 1/2 × 2 × 100 × 100

                        = 10000joules

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