Question

10. A skateboarder with a mass of 30 kg ends up with a final velocity of 7 m/s after a

push of 360 joules of work. What was the skateboarders initial velocity?

Answer 1

5 m/s

Step-by-step explanation:

The net work is equal to change in kinetic energy hence

`360J= 0.5m(v_f^(2)-v_i^(2))` where m is the mass of the skateboarder, v is the speed and the subscripts i and v represent initial and final respectively. Substituting 7 m/s for final velocity and 30 for mass m then

`360=0.5* 30* (7^(2)-v_i^(2))\\360=15(49-v_i^(2))\\\frac {360}{15}=49-v_1^(2)\\v_i^(2)=49-24\\v_i=√(25)\\v_i=5 m/s`