Question

# A 51.7-kg skateboarder starts out with a speed of 1.67 m/s. He does 87.4 J of...

A 51.7-kg skateboarder starts out with a speed of 1.67 m/s. He does 87.4 J of work on himself by pushing with his feet against the ground. In addition, friction does -258 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 6.99 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.

Solution) m = 51.7 kg

u = 1.67 m/s

W1 = 87.4 J

W2 = - 258 J

v = 6.99 m/s

(a) PEf - PEo = ?

We have

W = (KEf - KEo) + (PEf - PEo)

W = W1 + W2 = 87.4 - 258 = - 170.6 J

W = - 170.6 J

KEf = (1/2)(m)(v^2) = (1/2)(51.7)(6.99^2) =  1263.03 J

KEo = (1/2)(m)(u^2) = (1/2)(51.7)(1.67^2) = 72.1 J

PEf - PEo = W - (KEf - KEo) = - 170.6 - 1263.03 + 72.1

PEf - PEo = - 1361.53 J

(b) h = ?

PEf - PEo = mgh

- 1361.53 = 51.7×9.8×h

h = - (1361.53)/(51.7×9.8)

h = - 2.68 m

Therefore h = 2.68 m below which is indicated by negative sign

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