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lisabon 2012 [21]
3 years ago
12

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 th

e 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.
Physics
1 answer:
n200080 [17]3 years ago
6 0

Answer:

a. Wgra=786.09J

b. 1.28m

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=\frac{m}{2}(v_f^2-v_0^2)\\W_{tot}=\frac{62.6kg}{2}((6.55\frac{m}{s})^2-(2.18\frac{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=\frac{W_{grav}}{mg}=\frac{786.09J}{(62.6kg)(9.8\frac{m}{s^2})}=1.28m

HOPE THIS HELPS!!

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