The velocity of a body of mass 10kg increases from 4m/s to 8m/s when a force acts on it for 2s. (a) What is the momentum before and after the force acts?(c) What is the value of the force?
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Answered by
3
Force is 20N
U=4m/s
V=8m/s
T=2s
V=u+at
8=4+a*3
a=2
F=ma
F=10*2
F=20N
U=4m/s
V=8m/s
T=2s
V=u+at
8=4+a*3
a=2
F=ma
F=10*2
F=20N
Answered by
4
given, mass m=10kg
initial velocity u=4m/s
final velocity V=8m/s
time t=2s
therefore force F=change of momentum
F=m(v-u)/t
F=10(8-4)N
F=40/2N
F=20N
momentum p=mu (before the force act)
p=10x4
=40kgm/s
momentum p= m(8-4) ( after the force act ]
p=10x4
=40kgm/s
initial velocity u=4m/s
final velocity V=8m/s
time t=2s
therefore force F=change of momentum
F=m(v-u)/t
F=10(8-4)N
F=40/2N
F=20N
momentum p=mu (before the force act)
p=10x4
=40kgm/s
momentum p= m(8-4) ( after the force act ]
p=10x4
=40kgm/s
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