A bullet of mass 0.04kg moving with a speed of 90 ms^-1 enters a heavy wooden block and is stopped after a distance of 60cm.What is the average resistive force excerted by the block on the bullet?
The retardation 'a' of the bullet (assumed constant) is given by
a=-u^2/2s=-90*90/2*0.6ms^-2=-6750ms^-2
Hello Student,
The wooden block provides a constant resistive force and hence the bullet deccelerates at a constant rate, let it be a .
Since the bullet stops after 60cm there fore final velocity is 0
1. v² - u² = 2as
or a = -u²/2a = 90²/2*0.6 = -6750m/s²
2.Therefore the retarding force will be
Mass of bullet * decceleration = 0.04kg * -6750m/s² = -270N
Hello,
This is a simple question and can be solved with Newton's Second Law.
According to it, F=ma.
here m is mass of bullet. (0.04kg)
a is acceleration of bullet ( can be calculated using speed and distance)
we know, v^2-u^2=2as
here, v=0 m/s , u=90m/s and s is 0.6m
therefore, a= - 6750m/s^2
Hence we can find resistive force (F) = 0.04 x ( - 6750) = - 270N.
Here negative sign indicates that the force applied by the block is in opposite direction to the velocity of the bullet.
Hope this helps.
Thanks
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