Given :
A hockey puck with a mass of 0.18 kg is at rest on the horizontal frictionless surface of the rink.
A player applies a horizontal force of 0.5 N to the puck.
To Find :
The speed and the traveled distance 5s later.
Solution :
Acceleration is :
a = F/m
a = 0.5/0.18 = 2.78 m/s² .
Now, velocity after 5 seconds is given by :
v = 0 + 2.78×5 m/s
v = 13.9 m/s
Also, distance travelled :
[tex]d = \dfrac{v^2-u^2}{2a}\\\\d = \dfrac{13.9^2-0^2}{2\times 2.78}\\\\d = 34.75 \ m[/tex]
Therefore, speed and the traveled distance 5 s later is 13.9 m/s and 34.75 m.
