A worker drives a 0.554 kg spike into a rail tie with a 2.30 kg sledgehammer. The hammer hits the spike with a speed of 62.7 m/s. If one fourth of the hammer’s kinetic energy is converted to the internal energy of the hammer and spike, how much does the total internal energy increase? Answer in units of J.

Respuesta :

Answer:

Increase in total energy will be equal to the increase in the internal energy i.e [tex]1130.246[/tex] Joules

Explanation:

Given

Weight of sledge hammer [tex]= 2.30[/tex] kilogram

Speed of sledge hammer [tex]= 62.7[/tex] meter per second

Kinetic energy is equal to half the product of mass and velocity square

[tex]K E = \frac{1}{2} mv^2[/tex]

Substituting the value of mass and velocity, we get -

[tex]KE = 0.5 * 2.30 * 62.7^2\\KE = 4520.9835[/tex]

It is given that one fourth of the energy is converted into internal energy

One fourth of kinetic energy is equal to

[tex]\frac{1}{4} * 4520.9835\\= 1130.246[/tex]

Increase in total energy will be equal to the increase in the internal energy i.e [tex]1130.246[/tex] Joules

Answer:

The total internal increases energy is 1506.9 J.

Explanation:

Given that,

Mass of spike = 0.554 kg

Mass of sledgehammer.= 2.30 kg

Speed = 62.7 m/s

One-third of the kinetic energy of the hammer is converted into the internal energy.

We need to calculate the total internal energy increase

Using given relation of kinetic energy and internal energy

[tex]E=\dfrac{K.E}{3}[/tex]

Where, E = internal energy

K.E = kinetic energy

Put the value into the formula

[tex]E=\dfrac{1}{2\times3}\times2.30\times(62.7)^2[/tex]

[tex]E=1506.9\ J[/tex]

Hence, The total internal increases energy is 1506.9 J.