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
[tex]m_{1}u_{1}=m_{2}v_{2}[/tex]
[tex]80\times10^{-3}\times80=8.0\times v[/tex]
[tex]v =\dfrac{80\times10^{-3}\times80}{8.0}[/tex]
[tex]v = 0.8\ m/s[/tex]
The work done is equal to the change in kinetic energy
[tex]W=\Delta KE[/tex]
[tex]W=\dfrac{1}{2}mv^2[/tex]
[tex]W=\dfrac{1}{2}\times8.0\times0.8^2[/tex]
[tex]W=2.56\ J[/tex]
We know that,
The work is defined as,
[tex]W = F\cdot d[/tex]
[tex]d = \dfrac{W}{F}[/tex]
[tex]d=\dfrac{2.56}{7.1}[/tex]
[tex]d =0.36\ m[/tex]
Hence, The block will be slide 0.36 m on the ice.
