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2 years ago

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A 55.6-kg skateboarder starts out with a speed of 2.44 m/s. He does 80.4 J of work on himself by pushing with his feet against t

he ground. In addition, friction does -244 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 7.24 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.

1 answer:

Vikentia [17]2 years ago

7 0

(a) -1620.8 J

The initial kinetic energy of the skateboarder is:

$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(55.6 kg)(2.44 m/s)^2=165.5 J$

where m is the skateboarder's mass and u his initial speed;

While the final kinetic energy is

$K_f = \frac{1}{2}mv^2 = \frac{1}{2}(55.6 kg)(7.24 m/s)^2=1457.2 J$

where v is his final speed.

So the change in kinetic energy is

$\Delta K=K_f - K_i = 1457.2 J -165.6 J = 1291.6 J$

According to the work-energy theorem, the change in mechanical energy (kinetic+potential) of the skateboarder is equal to the work done on it:

$\Delta K + \Delta E_p = W + W_f$

where

$W = 80.4 J$ is the work done by the skateboarder on himself

$W_f = -244 J$ is the work done by friction

$\Delta E_p = E_p_f - E_p_i$ is the change in gravitational potential energy

Solving for $\Delta E_p$ ,

$\Delta E_p = W+W_f - \Delta K=80.4 J - 244 J - 1457.2 J = -1620.8 J$

(b) 2.97 m

The change in potential energy of the skateboarder can be written as

$\Delta E_p = mg \Delta h$

where

m = 55.6 kg is the mass

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h$ is the change in vertical height of the skateboarder

Solving for $\Delta h$ ,

$\Delta h = \frac{\Delta E_p}{mg}=\frac{-1620.8 J}{(55.6 kg)(9.8 m/s^2)}=-2.97 m$

Where the negative sign means the skateboarder has moved downwards. Since we are interested only in the absolute value, the answer is

h = 2.97 m

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