Question

a force of 30N acts on a body of mass 2.0 kg starting force from rest up to distance of 3.0m . then the force reduece to 15 N and acts in the same direction up to 2.0m calculate the final kinetic energy of the body

Answer 1

E=mc^2 pls check. .....

Answer 2

Given,

m=2kg

F=30N

Distance covered  (S)= 3m

From   F = ma   relation

a= F/m

=30/2

=15m/s²

From equation of motion

v² = u² + 2aS

v² = (0)² + 2 * 15* 3 (Since motion of body starts from rest)

 = 90

In second case,

the final velocity produced by first force becomes the initial velocity of the next force,such that

v=u

u² = v² = 90

Given ,

Force F= 15N

S = 2m

From F = ma relation

a= F/m

= 15/2

= 7.5m/s²

From equation of motion

v² = u² + 2aS

v² = 90 + 2 * 7.5* 2 (Since u² = v²  )

= 90 + 30

 = 120

Final kinetic energy = 1/2 * mass of moving body * (final velocity)²

=  1/2 * 2 * 120

= 120 joules (Ans)

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