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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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Answer» the BULLET (m) = 0.03 kgInitial velocity of the bullet (u) = 400 m/sFinal velocity of the bullet (V) = 0 m/sDistance travelled by the bullet = 12 CM = 0.12 mForce (F) = ?Force is given by,F = ma ................(1)v = u + at ................(2)Putting the given values in eq(2) we get,0 = 400 + a×0.0003a = - 1.3 × 10^6 m/sec^2From eq(1),F = 0.03 KG × 1.3 × 106 m/sec^2F = 39 × 10^3 N = 39000 N |
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