Question

Using a 2.44kg sledge hammer, a railroad worker drives a 0.55kg spike into a railroad tie. The sledge hits the spike with a speed of 62 m/s. If 0.33 of the sledge’s kinetic energy is converted to the internal energy of the sedge and spike, how much does the total internal energy increase?

Answer 1

Answer:

1547.6 J

Explanation:

From the law of conservation of energy, the sum of initial and final energy is the same. Energy can only be transformed into other forms but can't be created nor lost

Initial kinetic energy of the hammer system is given by

$KE_i=0.5mv^{2}$ where m is mass of the body and v is the velocity

Substituting m for mass of hammer which is 2.44 Kg and v for 62 m/s

$KE_i=0.5*2.44 Kg*(62 m/s)^{2}=4689.68 J$

Since the hammer comes to rest, the final kinetic energy is zero since it's velocity is zero

$KE_f=0 J$

Change in kinetic energy is given by

$\triangle KE= KE_f-KE_i$ hence 0-4689.68  J=-4689.68  J

Change in total internal energy is given by

$\triangle PE-\triangle KE$ and in this case the change in potential energy is zero

$\triangle PE-\triangle KE=0--4689.68 J=4689.68 J$

Since only 0.33 of this energy is converted hence

$0.33\times4689.68 J=1547.5944 J\approx 1547.6 J$