Question

A block of mass 2kg is kept on a rough horizontal surface with coefficient of friction 0.3 .A force of 10N is applied in the rightward direction for 5sec.find(a)Work done by the applied force in 5sec,(b) work done by friction in 5sec,(c)Net workdone on the body in 5sec,(d)velocity of the body after 5sec and after 10 sec

Answer 1

acceleration=ubalanced force/total mass
a=10-0.3×10×2/2(frictional force=mew×normal reaction;N=mg)
=10-6/2
=4/2=2
therefore a=2

a)work done=force.displacement
=ma.s(acceleration=velocity/time,2=v/5,v=10)
(displacement=velocity.time,10×5=50)
W=2×2×50
W=200

b)W=F.S
=6×50
=300

c)W=net force.s
=4×50
=200

d)after 5sec
v=u+at
v=10+2×5
v=20
after 10 sec.
v=10+2×10
v=30