Question

A hammer of mass 1 kg moving at 25m/s strikes a nail . The nail stops the hammer in a very short time of 0.02 seconds . What is the force of the nail on the hammer ? ( Please don't give unnecessary answer ) ​

Answer 1

Answer :-

Force of the nail on the hammer is a retarding force of 1250 Newtons .

Explanation :-

We have :-

→ Mass of the hammer (m) = 1 kg

→ Initial velocity (u) = 25 m/s

→ Time taken (t) = 0.02 sec

→ Final velocity (v) = 0 m/s

[As the hammer finally stops ] .

________________________________

Firstly, let's calculate the acceleration of the hammer by using the 1st equation of motion .

v = u + at

⇒ 0 = 25 + a(0.02)

⇒ 0 - 25 = 0.02a

⇒ 0.02a = -25

⇒ a = -25/0.02

⇒ a = -1250 m/s²

Now, we can calculate the force by using Newton's 2nd law of motion .

F = ma

⇒ F = 1(-1250)

⇒ F = -1250 N

[Here, -ve sign represents reatrding force].

Answer 2

Answer:

Given :-

• A hammer of mass 1 kg moving at 25 m/s strikes a nail.
• The nail stops the hammer in a very short time of 0.02 seconds.

To Find :-

• What is the force of the nail on the hammer.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

$\clubsuit$ Force Formula :

$\mapsto \sf\boxed{\bold{\pink{F =\: ma}}}$

Solution :-

First, we have to find the acceleration :

Given :

$\bigstar\: \: \bf{Final\: Velocity\: (v) =\: 0\: m/s}$

$\bigstar\: \: \bf{Initial\: Velocity\: (u) =\: 25\: m/s}$

$\bigstar\: \: \bf{Time\: (t) =\: 0.02\: seconds}$

According to the question by using the formula we get,

$\implies \sf 0 =\: 25 + a(0.02)$

$\implies \sf 0 =\: 25 + 0.02a$

$\implies \sf 0 - 25 =\: 0.02a$

$\implies \sf - 25 =\: 0.02a$

$\implies \sf \dfrac{- 25}{0.02} =\: a$

$\implies \sf \dfrac{- 25}{\dfrac{2}{100}} =\: a$

$\implies \sf \dfrac{- 25}{1} \times \dfrac{100}{2} =\: a$

$\implies \sf \dfrac{- \cancel{2500}}{\cancel{2}} =\: a$

$\implies \sf \dfrac{- 1250}{1} =\: a$

$\implies \sf - 1250 =\: a$

$\implies \sf \bold{\purple{a =\: - 1250\: m/s^2}}$

Hence, the acceleration is - 1250 m/s².

Now, we have to find the force of the nail on the hammer :

Given :

$\bigstar\: \: \bf{Mass\: (m) =\: 1\: kg}$

$\bigstar\: \: \bf{Acceleration\: (a) =\: - 1250\: m/s^2}$

According to the question by using the formula we get,

$\longrightarrow \sf F =\: (1)(- 1250)$

$\longrightarrow \sf F =\: 1 \times (- 1250)$

$\longrightarrow \sf\bold{\red{F =\: - 1250\: N}}$

${\small{\bold{\underline{\therefore\: The\: force\: of\: the\: nail\: on\: the\: hammer\: is\: -\: 1250\: N\: .}}}}$