Problem 138 - A Complete Resource Book in Physics for JEE Main

Question

A block of mass \(5 \mathrm{~kg}\) is kept on a rough horizontal floor. It's given velocity is \(33 \mathrm{~m} / \mathrm{s}\) towards right. A force of \(20 \sqrt{2} \mathrm{~N}\) continuously acts on the block as shown. If the co- efficient of friction between block and floor is \(0.5\), find the velocity of the block after 5 seconds \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\).

Step-by-Step Solution

Answer

The velocity of the block after 5 seconds is \(38 m/s\).

Step 1: Calculate the horizontal and vertical components of the force.

First, let's find the horizontal and vertical components of the external force that acts on the block. Given, the force is \(20\sqrt{2} N\). The force is acting at a 45-degree angle to the floor. Using trigonometry, the horizontal and vertical components of the force can be found as follows: Horizontal component: \(F_x = F \cdot \cos(45°) = F \cdot \frac{1}{\sqrt{2}}\) Vertical component: \(F_y = F \cdot \sin(45°) = F \cdot \frac{1}{\sqrt{2}}\) Substitute the value of \(F=20\sqrt{2} N\): \(F_x = 20 N\) \(F_y = 20 N\)

Step 2: Calculate the frictional force acting on the block.

Since there is friction between the block and the floor, we need to consider the frictional force acting on the block. The frictional force can be calculated using the formula: \(f = \mu \cdot N\) Where \(f\) is the frictional force, \(\mu\) is the coefficient of friction, and \(N\) is the normal force. In this problem, we know that \(\mu = 0.5\). Now, let's find the normal force. The normal force is equal to the gravitational force acting on the block minus the vertical component of external force: \(N = m \cdot g - F_y\) Substitute the values of \(m = 5 kg\), \(g = 10 m/s^2\), and \(F_y = 20 N\): \(N = 5 \cdot 10 - 20 = 30 N\) Now we can calculate the frictional force: \(f = 0.5 \cdot 30 = 15 N\)

Step 3: Calculate the net force acting on the block.

Next, we need to find the net force acting on the block along the horizontal axis. The net force is the difference between the horizontal component of the external force and the frictional force: \(F_{net} = F_x - f\) Substitute the values of \(F_x = 20 N\) and \(f = 15 N\): \(F_{net} = 20 - 15 = 5 N\)

Step 4: Calculate the acceleration of the block.

Now that we have the net force acting on the block, we can calculate the acceleration of the block using Newton's second law of motion: \(F_{net} = m \cdot a\) Where \(a\) is the acceleration. Rearrange the equation to find a: \(a = \frac{F_{net}}{m}\) Substitute the values of \(F_{net} = 5 N\) and \(m = 5 kg\): \(a = \frac{5}{5} = 1 m/s^2\)

Step 5: Calculate the final velocity of the block after 5 seconds.

Finally, we can calculate the velocity of the block after 5 seconds using the kinematic equation: \(v_f = v_i + a \cdot t\) Where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, and \(t\) is the time period. Substitute the values of \(v_i = 33 m/s\), \(a = 1 m/s^2\), and \(t = 5 s\): \(v_f = 33 + 1 \cdot 5 = 33 + 5 = 38 m/s\) The velocity of the block after 5 seconds is \(38 m/s\).