Question

A bullet of mass 0.05 kg moving with a speed of 80 m/s enters a heavy wooden block and is stopped after a distance of 64 cm. The average force exerted by the block is
a) 250 N
b) 150N
c) 200N
d) 220N​

Answer 1

Answer:

given mass of bullet = 0.05 kg

initial velocity = 80 m/ s

final velocity. = o

distance = 64 cm

= 0.64 m

we know

v^2=u^2-2as

0 = u^2 -2as

a = 80×80 /2×0.64

a = 5000 m/ s

F = ma

F = 0.05× 5000 = 250 N

Answer 2

Given : Mass of bullet = 0.05 kg

              Initial velocity = 80 m/ s

              Final velocity. = o

              Distance travelled = 64 cm  = 0.64 m

To Find : Average force exerted by the block

Solution :

Mass of bullet = 0.05 kg

Initial velocity = 80 m/ s

Final velocity. = o

Distance travelled = 64 cm  = 0.64 m

From 3rd equation of motion,

$\[2as = {v^2} - {u^2}\]$

$\[2as = 0 - {u^2}\]$

$\[a = - \frac{{{u^2}}}{{2s}}\]$

$\[a = - \frac{{80 \times 80}}{{2 \times 0.64}}\]$

a = -5000 m/$\[{s^2}\]$

The resistive force exerted on the block is  

f= ma

f= 0.05× -5000 = -250 N

Thus, The average force on the block is 250N.

Hence, The average force exerted by the block is 250N.