Question

A hammer of mass 400g move at 30m/sec strikes a nail.The nail stops the hammer in a very short time of 0.01sec. what is the force of the nail on the hammer.​

Answer 1

Answer:

1200 N

Explanation:

AS THE HAMMER GENERALLY IS ON REST IN HOUSES.

THEREFORE =>>>

INITIAL VELOCITY OF HAMMER =u = 0m/s

FINAL VELOCITY = v = 30m/s

IT IS GIVEN THAT NAIL STOPS THE HAMMER IN 0.01SECONDS , THUS,

TIME = 0.01 SECONDS.

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NOW FOR A BODY IN MOTION ACCELERATION IS THE RATIO OF DIFFERENCE OF IT'S FINAL VELOCITY AND INITIAL VELOCITY TO TIME TAKE TO REACH FINAL VELOCITY.

= v - u/t

AFTER INPUTTING THE UPPER VALUES IN THIS WE GET =>>>>

$acc = 30 - 0 \div 0.01$

$acc = 30 \div 0.01$

$acc = 3000m \div {s}^{2}$

THEREFORE ACCELERATION OF THE HAMMER IS CALCULATED TO BE 3000M/S^2.

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NOW, NEWTON'S SECOND LAW OF MOTION SAYS THAT FORCE APPLIED ON A BODY IS EQUAL TO PRODUCT OF IT'S MASS AND ACCELERATION.

[ F = Ma ]

IN THIS CASE: FORCE APPLIED BY NAIL ON THE HAMMER =

F = 400g × 3000m

SINCE 400g = 0.4kg

THEREFORE

F = 0.4 × 3000m

= 1200N.

THUS FORCE APPLIED BY NAIL ON HAMMER IS 1200N.

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BY SCIVIBHANSHU

THANK YOU

STAY CURIOUS

Answer 2

$\LARGE{\underline{\underline{\sf{ANSWER:}}}}$

force is 1200 N

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Given:

A hammer of mass 400g move at 30m/sec strikes a nail.The nail stops the hammer in a very short time of 0.01sec. what is the force of the nail on the hammer.

Solution:

Mass , m = 0.4 kg

Change in velocity = initial velocity – final velocity

∆V = 30 – 0 = 30 ms –¹

Force , F = $\frac{m∆V}{t}\:=\:\frac{0.4\:×\:30}{0.01}\:=\:1200 N$

Hence , force is 1200 N