Question

a body of mass 10 kg is raised from certain depth if work done in raising it by 10 metre is 1600Joule in velocity at this position will be equal to g= 10 metre per second ​

Answer 1

so here work is done by two forces one is external Force for which work is given equals to 1600 Joule and another is gravitational force which has magnitude of of work

=> force × displacement

100 × (-10 )

-1000 joules

so total work done on body = 1600- 1000

600 Joule

on applying work energy theorem

work done by all forces = ∆ kinetic energy

600 =1/2× 10 v^2

v^2 = 120

v= 2√30 m/sec

hope it will help you

Answer 2

Answer:

Answer:

v=(20×2^1/2)/5

Explanation:

Here,

w=F×D

w=m×a×d

w=m×v×t^-1×d

1600=10×v×t^-1×10

16=v×t^-1

16=d/t×1/t

16=10/t^2

t^2=10/16

t=(5/8)^1/2

t=(5/2^3)^1/2

v=10/(5/2^3)^1/2

v=(20×2^1/2)/5

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