a bullet of mass 0.04 kg moving with velocity of 90m/sec enters a haevy woodn block and is stopped after a distance of 60cm what is average resistive force exerted by block on bullet
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use kinematic formula,
v^2=u^2+2as
after stopped v=0
0=(90)^2+2 x a x 60 x 10^-2
-8100 =0.6 a
a = -8100/0.6 = - 13500 m/s^2
hence,
resistive force = mass x acceleration
=0.04 x 13500 =540 N
v^2=u^2+2as
after stopped v=0
0=(90)^2+2 x a x 60 x 10^-2
-8100 =0.6 a
a = -8100/0.6 = - 13500 m/s^2
hence,
resistive force = mass x acceleration
=0.04 x 13500 =540 N
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