a body of mass 5 kg initially at rest is more by a horizontal force of 2 Newton on a smooth horizontal surface the work done by the force in 10 second is
Answers
Answered by
150
a=F/m
a=2/5 m/s²
s=1/2at² ( because u is zero)
s=1/2*2/5*10²
s=2/10*100
s=20m
w=F.S
w=2*20
w=40 J
a=2/5 m/s²
s=1/2at² ( because u is zero)
s=1/2*2/5*10²
s=2/10*100
s=20m
w=F.S
w=2*20
w=40 J
Answered by
37
A group of mass 5 kg at first very still is more by an even power of 2 Newton on a smooth level surface the work done by the power in 10 second is
A = F/m a = 2/5 m/s² s=1/2at² (on the grounds that u is zero) s=1/2*2/5*10² s = 2/10*100 s = 20m w=F.S w=2*20 w=40 J.
Similar questions