Question

A bullet of mass 0.04 kg moving with a speed of 90 m/s enters a heavy wooden block and stopped after a distance of 0.6m. What is the average resistive force exerted by the block on the?

Answer 1

$\huge\mathfrak{Answer:-}$

$m_{B}$$= 0.04 \: kg$

$u = 90 \: m/s$

$s = 0.6 \: m$

$\bold{We\:know\:that,}$

$v ^{2} = u ^{2} + 2as$

$0 = (90)^{2} + 2 \times a \times 0.6$

$0 = 8100 + 2 \times a \times 0.6$

$0 = 8100 + 1.2 \times a$

$- 1.2a = 8100$

$a = - \frac{8100}{1.2}$

$a = - 6750\:m/s^{2}$

$\bold{Therefore,}$

Resistive force

$= m \times a$

$= 0.04 \times 6750$

$= 270N$

$\text{Be\:Brainly}$❤️

Answer 2

mB​ = 0.04 kg

u = 90 m/s

s = 0.6 m



$v ^{2} = u ^{2} + 2as$

$\bf{0 = (90)^{2} + 2 \times a \times 0.60}$

$\bf{0 = 8100 + 2 \times a \times 0.60}$

0=8100+1.2×a 

−1.2a=8100 

$\bf{a = - \frac{8100}{1.2}}$

$\bf{a = - 6750\:m/s^{2}}$



Resistive force 

=m×a 

=0.04×6750 

= 270N

Thanks