Question

On the horizontal surface of a truck \((=0.6)\), a block of mass $1 \mathrm{~kg}\( is placed. If the truck is accelerating at the rate of \)5 \mathrm{~m} / \mathrm{s}^{2}$ then frictional force on the block will be \(\mathrm{N}\) (A) 5 (B) 6 (C) \(5.88\) (D) 8

Short answer

The frictional force on the block is \(5.88\mathrm{N}\).

Step-by-step solution

Find the normal force on the block

To find the normal force on the block, we first need to find the weight of the block, which is the gravitational force acting on it: \[W = mg\] where \(W\) = weight of the block, \(m\) = mass of the block = \(1\mathrm{~kg}\), and \(g\) = acceleration due to gravity = \(9.8\mathrm{~m} / \mathrm{s}^{2}\). Substituting the values and calculating the weight: \[W = (1\mathrm{~kg})(9.8\mathrm{~m} / \mathrm{s}^{2}) = 9.8\mathrm{N}\] Since the block is resting on a horizontal surface and there is no vertical acceleration, the normal force (\(N\)) is equal to the weight of the block: \[N = W = 9.8\mathrm{N}\]

Calculate the frictional force

To find the frictional force (\(f\)), we use the frictional force equation: \[f = μN\] where \(f\) = frictional force, \(μ\) = frictional coefficient = \(0.6\), and \(N\) = normal force = \(9.8\mathrm{N}\). Substituting the values and calculating the frictional force: \[f = (0.6)(9.8\mathrm{N}) = 5.88\mathrm{N}\] Hence, the frictional force on the block is \(5.88\mathrm{N}\). The correct answer is (C).