A forging hammer of mass m_1 = 1.2 t strikes a piece of iron…

A forging hammer of mass $m_1 = 1.2\ \mathrm{t}$ strikes a piece of iron lying on an anvil of mass $m_2 = 25\ \mathrm{t}$ at $v_1 = 5\ \mathrm{m/s}$. (a) How much energy goes into deforming the iron? (b) What is the efficiency of the blow?

Explanation of Method

Efficiency of a forging hammer.

Step 1: (a) Deformation energy

After the blow, hammer and anvil move together (inelastic impact): $v_2 = \frac{m_1 v_1}{m_1 + m_2}$.

\[ E_{\text{def}} = \frac{m_1 v_1^2}{2} - \frac{(m_1 + m_2)v_2^2}{2}, \]

\[ E_{\text{def}} = \frac{m_1 v_1^2}{2} \cdot \frac{m_2}{m_1 + m_2}. \]

\[E_{\text{def}} = 15\ \mathrm{kJ} \cdot \frac{25}{26.2} \approx 14.3\ \mathrm{kJ}\]

Step 2: (b) Efficiency

In forging, the useful energy is the deformation energy: \[\eta = \frac{E_{\text{def}}}{E_0} = \frac{m_2}{m_1 + m_2} \approx 0.954 = 95\%\]

$E_0 = 15\ \mathrm{kJ}$ $\text{hammer}$ $\text{deformation } 14.3\ \mathrm{kJ}$ $\text{after the blow}$ $0.69$ $\eta = \frac{m_2}{m_1 + m_2} \approx 95\%$

For forging, the useful energy is the part that deforms the workpiece. With a heavy anvil almost all of the hammer's energy goes there; only 4.6% is left as motion of hammer and anvil.

Result

\[\boxed{\,E_{\text{def}} = 15\ \mathrm{kJ} \cdot \frac{25}{26.2} \approx 14.3\ \mathrm{kJ}\,}\]

\[\boxed{\,\eta = \frac{E_{\text{def}}}{E_0} = \frac{m_2}{m_1 + m_2} \approx 0.954 = 95\%\,}\]

Remark

Check. The original results are right (14.3 kJ is more precise than its 14.2). Its formula for the efficiency, $\eta = 1 - E_{\text{def}}/E_0$, is wrong: that would give $m_1/(m_1 + m_2) \approx 4.6\%$. It is $\eta = E_{\text{def}}/E_0$. (For pile-driving, where the useful part is the motion, the opposite, a light anvil, would be better.)