Question

Position \( x \) of the particle of mass 2 kg varies as function of time as \( x = t^2 + t + 1 \). Find out work done on the particle from \( t_1 = 2 \) sec to \( t_2 = 3 \) sec.

• A. 18 joule
• B. 30 joule
• C. 34 joule
• D. 24 joule

Hint:

Work done on a particle is equal to the change in its kinetic energy. The velocity of the particle can be found by differentiating its position with respect to time.

Answer 1

The Correct Option is D

Step 1: Understand the formula for work done.
The work done on a particle is given by the change in kinetic energy: \[ W = \Delta K.E. = \frac{1}{2} m \left( v_2^2 - v_1^2 \right) \] where \( m \) is the mass of the particle, and \( v_1 \) and \( v_2 \) are the initial and final velocities of the particle.
Step 2: Find the velocity of the particle.
The position \( x \) of the particle is given by: \[ x = t^2 + t + 1 \] To find the velocity, we differentiate \( x \) with respect to time \( t \): \[ v = \frac{dx}{dt} = 2t + 1 \]
Step 3: Calculate the velocities at \( t_1 = 2 \) sec and \( t_2 = 3 \) sec.
At \( t_1 = 2 \) sec: \[ v_1 = 2(2) + 1 = 5 \, \text{m/s} \] At \( t_2 = 3 \) sec: \[ v_2 = 2(3) + 1 = 7 \, \text{m/s} \]
Step 4: Calculate the work done.
The mass of the particle is given as \( m = 2 \, \text{kg} \). Using the work-energy theorem: \[ W = \frac{1}{2} \times 2 \left( 7^2 - 5^2 \right) \] \[ W = (49 - 25) = 24 \, \text{joules} \] Thus, the work done on the particle is 24 joules.