Question

The man fires an 80 g arrow so that it is moving at 80 m/s when it hits and embeds in a 8.0 kg block resting on ice. How far will the block slide on the ice before stopping? A 7.1 N friction force opposes its motion.

Answer 1

Answer:

The block will be slide 0.36 m on the ice.

Explanation:

Given that,

Mass of arrow m₁= 80 g

Velocity of arrow u₁= 80 m/s

Mass of block m₂= 8.0 kg

Force F = 7.1 N

Using conservation of momentum

$m_{1}u_{1}=m_{2}v_{2}$

$80\times10^{-3}\times80=8.0\times v$

$v =\dfrac{80\times10^{-3}\times80}{8.0}$

$v = 0.8\ m/s$

The work done is equal to the change in kinetic energy

$W=\Delta KE$

$W=\dfrac{1}{2}mv^2$

$W=\dfrac{1}{2}\times8.0\times0.8^2$

$W=2.56\ J$

We know that,

The work is defined as,

$W = F\cdot d$

$d = \dfrac{W}{F}$

$d=\dfrac{2.56}{7.1}$

$d =0.36\ m$

Hence, The block will be slide 0.36 m on the ice.