## A 0.77 kg spike is hammered into a railroad tie. The initial speed of the spike is equal to 3.5 m/s. If

Question

A 0.77 kg spike is hammered into a railroad

tie. The initial speed of the spike is equal to

3.5 m/s.

If the tie and spike together absorb 88.3

percent of the spike’s initial kinetic energy

as internal energy, calculate the increase in

internal energy of the tie and spike.

in progress 0
2 months 2021-07-31T19:41:18+00:00 2 Answers 4 views 0

∆E = 4.16J

Change in internal energy of the tie and spike is 4.16J

Explanation:

Given;

Mass of spike m = 0.77kg

Initial Speed of spike v = 3.5m/s

Efficiency = 88.3%

Kinetic energy of spike can be calculated as;

K.E = 1/2 × mv^2

Substituting the values;

K.E = 1/2 × 0.77 × 3.5^2

K.E = 4.71625J

Change in internal energy of the tie and spike;

∆E = 88.3% of K.E

∆E = 0.883 × 4.71625J

∆E = 4.16J

Change in internal energy of the tie and spike is 4.16J

4.168  J

Explanation:

Applying the formula of kinetic energy,

Ek = 1/2mv²………………… Equation 1

Where Ek = Kinetic energy of the spike, m = mass of the spike, v = initial velocity of the spike

Given: m = 0.77 kg, v = 3.5 m/s.

Substitute into equation 1

Ek = 1/2(0.77)(3.5²)

Ek = 4.72 J.

From the question, if the tie and the spike together absorb, 88.3 percent of the spike’s initial kinetic energy.

Ek’ = 88.3/100(4.72)

Ek’ = 4.168  J