Question

A 62.6-kg skateboarder starts out with a speed of 2.18 m/s. He does 113 J of work on himself by pushing with his feet against the ground. In addition, friction does -295 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 6.55 m/s. (a) Calculate the change (PEf - PE0) in the gravitational potential energy. (b) How much has the vertical height of the skater changed? Give the absolute value.

Answer 1

a. Wgra=786.09J

b. 1.28m

Step-by-step explanation:

The change in the potential energy is the work done by the gravitational force.

For this problem you have to take into account that the total work done is given by the change in the kinetic energy

`W_(tot)=\Delta E_k=(m)/(2)(v_f^2-v_0^2)\\W_(tot)=(62.6kg)/(2)((6.55(m)/(s))^2-(2.18(m)/(s))^2)=1194.09J`

Furthermore the total work is the contribution of the work done by the skater, the gravitational force and the friction

`W_(tot)=W_(ska)+W_(fric)+W_(gra)`

(a) by separating Wfric you have

`W_(gra)=W_(tot)-W_(fric)-W_(ska)=1194.09J-295J-113J=786.09J`

(b) It is only necessary to use the expression for the work done by gravitational force

`W_(grav)=mgh\\h=(W_(grav))/(mg)=(786.09J)/((62.6kg)(9.8(m)/(s^2)))=1.28m`

HOPE THIS HELPS!!