A 0.71 kg spike is hammered into a railroad tie. The initial speed of the spike is equal to 3.8 m/s. If the tie and spike together absorb 38.2 percent of the spike’s initial kinetic energy as internal energy, calculate the increase in internal energy of the tie and spike. Answer in units of J.

Respuesta :

Answer:

1.96 J

Explanation:

From the law of conservation of energy

ΔU + ΔK = 0  where ΔU = internal energy change and ΔK = kinetic energy change. We neglect potential energy change since we are not given any information about it.

ΔU = -ΔK

ΔK = K₂ - K₁ where  K₁ = initial kinetic energy and K₂ = final kinetic energy = 0 where  ΔU = 0.382K₁

     = 0.382mv²/2  where m = mass of spike = 0.71 kg and v = initial speed of spike = 3.8 m/s

     = 0.382 × 0.71 kg × (3.8 m/s)²/2

     = 1.96 J

When The increase in internal energy of the tie and spike is = 1.96 J Applying the law of conservation of energy.

What is the Law of conservation of energy?

The law of conservation of energy are:

Then, ΔU + ΔK = 0 where ΔU is = internal energy change and ΔK is = kinetic energy change. Then We neglect potential energy change since we are not given any information about it.

After that, ΔU is = -ΔK

Then, ΔK is = K₂ - K₁ where K₁ = initial kinetic energy and also K₂ is = final kinetic energy = 0 where, ΔU is = 0.382K₁

Then, = 0.382mv²/2 where m is = mass of spike = 0.71 kg and v is = initial speed of spike is = 3.8 m/s

After that, = 0.382 × 0.71 kg × (3.8 m/s)²/2

Therefore, = 1.96 J

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