Work-energy Via Newton’s Laws

Question

A 2 kg block is pushed across a rough horizontal surface by a 16 N force. It moves at constant velocity. What is the coefficient of kinetic friction? (g = 10 m/s²)

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1. **Equilibrium Condition**: 'Constant velocity' means zero acceleration ($a=0$). By Newton's First Law, net force must be zero.
2. **Force Balance**:
   - Applied Force ($F$) $=$ Kinetic Friction ($f_k$)
   - $16$ N $= \mu_k N$.
3. **Normal Force**: On a horizontal surface, $N = mg = 2 	imes 10 = 20$ N.
4. **Solve for Coefficient**:
   - $16 = \mu_k 	imes 20$
   - $\mu_k = \frac{16}{20} = \frac{8}{10} = 0.80$.
5. **Result**: The coefficient of kinetic friction is **$0.80$**.

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Work-energy Via Newton’s Laws

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