Topic: Engineering Mechanics
A body of mass 10 kg is initially stationary on a $45^{\circ}$ inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.3. The body slides down the plane and attains a velocity of $20~m/s.$ The distance travelled (in meter) by the body along the plane is close to
- A. 4.8
- B. 98
- C. 41
- D. 9.8
Answer: C. 41
Explanation: From the FBD for mass m, equating the net forces along the plane: $ma = mg\sin 45^{\circ} - \mu(mg\cos 45^{\circ})$. $a = 9.81 \times \frac{1}{\sqrt{2}} \times [1 - 0.3] = 9.81 \times \frac{1}{\sqrt{2}} \times 0.7 = 4.85~m/s^{2}$. Using equation of motion $v^{2} = u^{2} + 2as \Rightarrow 20^{2} = 0^{2} + 2 \times 4.85 \times s \Rightarrow s = \frac{400}{9.7} = 41.2~m$. The closest option is 41.