Question

A bullet pf mass 0.04kg movimg with a speed of 90m/s inter having a wood blocks a is stopped after tranversing at a distance 6 cm .what is the average resistance force a exerted on the block of the object

Answer 1

A Bullet of mass 0.04 Kg moving with speed of 90 m/s inter having a wood block is stopped after covering distance 6 cm .
so, we have given,
mass of bullet , m = 0.04 kg
speed of bullet , u = 90 m/s
and distance covered by bullet in wood block , S = 6 cm
After penetration on wood block , bullet will be rest.
Means final velocity of bullet , v = 0
Use formula,
v² = u² + 2aS
0 = (90)² + 2a × 6/100
a = 8100 × 100/12 = - 6.75 × 10⁴ = -67500 m/s²

Hence, average resistance force exerted on the block , F = ma = -67500 × 0.04
= -2700 N [ negative sign indicates that force is resistive ]

Answer 2

Given :
mass  of bullet =m=0.04kg

initial speed=u=90m/s

Final velcoity=v= 0 m/s

Distance =s=6cm= 6/100 m


From third equations of motion:
v² - u²= 2as

0² -90² = 2 a x6/100

a= - 8100x100/12
= -67500 m/s2

Average resistance force = F= ma = 0.04 x -67500
=- 2700 N

Negative sign indicates it is the resistance  force