A block of mass $4kg$ is placed in contact with the front vertical surface of a lorry.The coefficient of friction between the vertical surface and block is $0.8$.The lorry is moving with an acceleration of $15m/s^2$.The force of friction between lorry and block is $\left(g=10m{s}^{-2}\right)$a. $48N$ b. $24N$ c. $40N$ d. Zero

Sure! Here are the step-by-step solutions for the given problem: We are given the following equations: Weight of an object: $W=mg$ Force of friction: $f=\mu ma$ We are also given the condition that the mass ($m$) is less than the force of friction ($f$). To solve the problem, we can use the following steps: Step 1: Write down the given equations. $W=mg$$f=\mu ma$ Step 2: Substitute the weight equation into the friction equation. $f=\mu ma⟹f=\mu m\cdot g$ Step 3: Rewrite the condition in terms of $f$ and $m$. $m<f⟹m<\mu m\cdot g$ Step 4: Solve for $g$. $m\frac{1}{\mu }$ Step 5: Write the final solution. $g>\frac{1}{\mu }$ Therefore, the minimum value of $g$ is $\frac{1}{\mu }$ in order for the given condition to be satisfied.