The initial kinetic energy of the bullet is:
\[ K_i = \frac{1}{2} m (100)^2 \]
The final kinetic energy of the bullet is:
\[ K_f = \frac{1}{2} m (40)^2 \]
The percentage loss in kinetic energy is given by:
\[ \% \text{loss} = \frac{|K_f - K_i|}{K_i} \times 100 \]
Substituting the expressions for \( K_i \) and \( K_f \):
\[ \% \text{loss} = \frac{\left| \frac{1}{2} m (40)^2 - \frac{1}{2} m (100)^2 \right|}{\frac{1}{2} m (100)^2} \times 100 \]
Simplify:
\[ \% \text{loss} = \frac{|1600 - 10000|}{10000} \times 100 \]
\[ \% \text{loss} = \frac{8400}{10000} \times 100 = 84\% \]
Final Answer: 84% (Option 4)
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).