Question

B. A hammer of mass 4 kg, falling from a height of 50 cm, hits a
nail, partially fixed on a surface, and stops in 0.1 s. Find the
force exerted on the nail in N.
[Ans. 125.22 N]​

Answer 1

Hey Dear,

◆ Answer -

F = 63.2 N

◆ Explaination -

Velocity of the falling hammer is calculated as -

u = √(2gh)

u = √(2 × 10 × 0.5)

u = √10

u = 3.16 m/s

When the hammer falls on the nail, it's decelerated until it stops.

v = u + at

0 = 3.16 + a × 0.1

a = -3.16/0.1

a = -31.6 m/s^2

Force exerted by nail on the hammer -

F = ma

F = 4 × (-31.6)

F = -63.2 N

Therefore, the force done is 63.2 N in the opposite direction.

Answer 2

Given :-

Mass of the hammer = 4 kg

Height the hammer fell = 50 cm

Time taken to stop =  0.1 sec

To Find :-

The  force exerted on the nail.

Solution :-

We know that,

• t = Time
• a = Acceleration
• u = Initial velocity
• f = Force
• m = Mass

Using the formula,

$\underline{\boxed{\sf Intial \ velocity=\sqrt{(2gh)} }}$

Substituting their values,

u = √(2 × 10 × 0.5)

u = √10

u = 3.16 m/s

Therefore, the initial velocity is 3.16 m/s.

Using first equation of motion,

$\underline{\boxed{\sf First \ equation \ of \ motion=v=u+at}}$

Given that,

Final velocity (v) = 0 (Since it comes to rest)

Initial velocity (u) = -3.16 m/s (Retardation since it falls from a height)

Time (t) = 0.1 sec

Substituting their values,

0 = 3.16 + a × 0.1

a = -3.16/0.1

a = -31.6 m/s²

By the formula,

$\underline{\boxed{\sf Force=Mass \times Acceleration}}$

Given that,

Mass (m) = 4 kg

Acceleration (a) = -31.6 m/s²

Substituting their values,

f = 4 × (-31.6)

f = -126.4 N

Therefore, the  force exerted on the nail is -126.4 N.