A ForCe of 500 Nacts on a body of mass 1000 kg and body is bought to rest within a distance of 6 um . Find the initial velocity and time taken by the body to come to rest.

To solve this problem, we'll determine both the initial velocity (\( v_0 \)) of the body and the time (\( t \)) it takes to come to rest under the action of a constant force. ### Given: - **Force (\( F \))**: 500 N - **Mass (\( m \))**: 1000 kg - **Distance (\( s \))**: 6 µm = \( 6 \times 10^{-6} \) meters ### 1. Finding the Initial Velocity (\( v_0 \)) We'll use the **work-energy principle**, which states that the work done by the force is equal to the change in kinetic energy of the body. \[ \text{Work Done} = F \times s = \frac{1}{2} m v_0^2 \] Solving for \( v_0 \): \[ v_0 = \sqrt{\frac{2 F s}{m}} \] Plugging in the values: \[ v_0 = \sqrt{\frac{2 \times 500 \, \text{N} \times 6 \times 10^{-6} \, \text{m}}{1000 \, \text{kg}}} \] \[ v_0 = \sqrt{\frac{6000 \times 10^{-6}}{1000}} = \sqrt{6 \times 10^{-6}} \approx 0.00245 \, \text{m/s} \, (2.45 \, \text{mm/s}) \] ### 2. Finding the Time Taken (\( t \)) First, determine the **deceleration** (\( a \)) caused by the force: \[ a = \frac{F}{m} = \frac{500 \, \text{N}}{1000 \, \text{kg}} = 0.5 \, \text{m/s}^2 \] Using the kinematic equation: \[ v = v_0 + a t \] Since the final velocity (\( v \)) is 0 (the body comes to rest): \[ 0 = v_0 - a t \implies t = \frac{v_0}{a} \] Plugging in the values: \[ t = \frac{0.00245 \, \text{m/s}}{0.5 \, \text{m/s}^2} = 0.0049 \, \text{seconds} \, (4.9 \, \text{milliseconds}) \] ### **Final Answer:** - **Initial Velocity (\( v_0 \))**: Approximately **2.45 mm/s** - **Time to Come to Rest (\( t \))**: Approximately **4.9 milliseconds**