A bullet of mass 25 gm and velocity 50 m/s passes through a board 7.5 cm thick and emerges with a velocity of 40m/s.what is the average force exerted by the boardon the bullet?
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Work = Fd where F is force, d is distance
But work is also W = ΔEk where Ek is kinetic energy
In general, kinetic energy Ek = ½mv^2
So here, ΔEk = ½m(v1^2 – v2^2)
= ½(0.025)(50^2 – 40^2) = 11.25J
Then Fd = F(0.075) = 11.25, so F = 11.25/0.075 = 150N
The average force is 150N.
But work is also W = ΔEk where Ek is kinetic energy
In general, kinetic energy Ek = ½mv^2
So here, ΔEk = ½m(v1^2 – v2^2)
= ½(0.025)(50^2 – 40^2) = 11.25J
Then Fd = F(0.075) = 11.25, so F = 11.25/0.075 = 150N
The average force is 150N.
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5
Answer:
150N
Explanation:
Applyng work-energy theorem
W=Chang in kinetic energy
_ f x s = 1/2 m ( v² -u²) by substituting
_ f x 0.075 = 1/2 (0.025) ( 40² -50²) solve for f
f = 150 N .
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