A bullet of mass 0.03kg travelling with speed of 400m/s penetrates 12cm into a fixed block of wood. calculate the average force exerted by the wood on the bullet
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Mass of the bullet (m) = 0.03 kg
Initial velocity of the bullet (u) = 400 m/s
Final velocity of the bullet (v) = 0 m/s
Distance travelled by the bullet = 12 cm
= 0.12 m
Force (F) = ?
Force is given by,
F = ma ................(1)
v = u + at ................(2)
Putting the given values in eq(2) we get,
0 = 400 + a×0.0003
a = - 1.3 × 10^6 m/sec^2
From eq(1),
F = 0.03 kg × 1.3 × 106 m/sec^2
F = 39 × 10^3 N = 39000 N
Initial velocity of the bullet (u) = 400 m/s
Final velocity of the bullet (v) = 0 m/s
Distance travelled by the bullet = 12 cm
= 0.12 m
Force (F) = ?
Force is given by,
F = ma ................(1)
v = u + at ................(2)
Putting the given values in eq(2) we get,
0 = 400 + a×0.0003
a = - 1.3 × 10^6 m/sec^2
From eq(1),
F = 0.03 kg × 1.3 × 106 m/sec^2
F = 39 × 10^3 N = 39000 N
ankit2842:
but how you have used the equation t is not given in the the question
you have used mass as time
sorry
V^2 = Vo^2 + 2a*d = 0.
400^2 + 2a*0.12 = 0,
160,000 + 0.24a = 0, a = -666,667 m/s^2.
F = M*a = 0.03 * (-666,667) = -20,000 N
400^2 + 2a*0.12 = 0,
160,000 + 0.24a = 0, a = -666,667 m/s^2.
F = M*a = 0.03 * (-666,667) = -20,000 N
I want to changed my answer
Well answered yrr....
Nyc...
Hii
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