Answer:
Acceleration=[tex]0.5m/s^2[/tex]
Speed=0.67 m/s
Explanation:
We are given that
Horizontal force=F=20 N
Mass of box=m=40 kg
We know that
Acceleration=[tex]a=\frac{F}{m}[/tex]
Using the formula
Acceleration of box=[tex]\frac{20}{40}=0.5m/s^2[/tex]
The acceleration of the box=[tex]0.5m/s^2[/tex]
Initial velocity=u=0
Force=F=30 N
Distance=s=0.3 m
[tex]a=\frac{30}{40}=\frac{3}{4} ms^{-2}[/tex]
[tex]v^2-u^2=2as[/tex]
Substitute the values
[tex]v^2-0=2\times \frac{3}{4}\times 0.3=0.45[/tex]
[tex]v^2=0.45[/tex]
[tex]v=\sqrt{0.45}=0.67m/s[/tex]
Hence, the speed of the box after it has been pulled a distance of 0.3 m=0.67 m/s