A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 0.12. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in ms⁻²$ms^{-2}$ is: (g = 10 m/s²)$(g = 10 \, \text{m/s}^2)$

Related Formula

Core Logic

Net force Fₙₑₜ = √(6² + 8²) = 10N$F_{\text{net}} = \sqrt{6^2 + 8^2} = 10\mathrm{N}$.

Acceleration a = (10)/(5) = 2 m/s²$a = \frac{10}{5} = 2 \mathrm{m/s}^2$. Angle with 8 N force: θ = (6)/(8) ⇒ θ = ⁻¹(3/4)$\tan \theta = \frac{6}{8} \Rightarrow \theta = \tan^{-1}(3/4)$ with 8 N force.

Pattern Recognition

Perpendicular force resultant vector acceleration.

Chapter Mix

Class 11 Physics: Laws of Motion