A person hammered a 0.5 kg something (substance ) with a hammer that its weight is 2.5 kg , with the velocity 65 m /s ,, if one third of the hammer's energy turn into the internal energy to both of the substance and the hammer , then what is the change (increasing ) internal energy ?
Thank you a lot
1/2 *2.5*65^2 * 1/3 =internal energy
To find the change in internal energy, we need to determine the initial energy and the energy transferred to the substance and the hammer.
1. Calculate the initial energy of the hammer:
Initial kinetic energy of the hammer (Ek) = 0.5 * mass * velocity^2
= 0.5 * 2.5 kg * (65 m/s)^2
2. Calculate the energy transferred to the substance and the hammer:
Energy transferred (Et) = 1/3 * Initial kinetic energy of the hammer
3. Determine the change in internal energy:
Change in internal energy (ΔE) = Energy transferred (Et)
To simplify the calculation, let's go through the steps:
1. Calculate the initial energy:
Ek = 0.5 * 2.5 kg * (65 m/s)^2
Ek = 0.5 * 2.5 kg * 4225 m^2/s^2
Ek = 5281.25 Joules (J)
2. Calculate the energy transferred:
Et = 1/3 * Ek
Et = 1/3 * 5281.25 J
Et = 1760.42 J
3. Determine the change in internal energy:
ΔE = Et
ΔE = 1760.42 J
Therefore, the change (increasing) internal energy is 1760.42 Joules.