Exams › NEET › Physics

A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by x = 3t − 4t² + t³, where x is in metres and t is in seconds. The work done during the first 4 seconds is

1. 576 mJ
2. 450 mJ
3. 490 mJ
4. 530 mJ

Correct answer: 576 mJ

Solution

The work done is calculated using the work-energy theorem, which states that the work done is equal to the change in kinetic energy. First, find the velocity as a function of time by differentiating the position function: v(t) = dx/dt = 3 - 8t + 3t². Then, calculate the initial and final velocities at t = 0 and t = 4. Use these velocities to find the initial and final kinetic energies, and compute the work done as the difference in kinetic energy. The result is 576 mJ.

Related NEET Physics questions

• Assertion A load when lifted vertically up requires greater effort than to roll the same load up the ramp. Reason The work done in lifting the load vertically up is less.
• The graph between the resistive force acting on a body and the position of the body travelling in a straight line is shown in the figure. The mass of the body is \( 25 \mathrm{kg} \) and initial velocity is 2 m/s.When the distance covered by the bodyi \( s 4 \mathrm{m}, \) its kinetic energy would be
• Potential energy increases with the increase in :
• Three different objects of masses m1, m2, and m3 are allowed to fall from rest and from the same point O along three different frictionless paths. The speeds of the three objects on reaching the ground will be in the ratio of
• What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?
• A particle of mass m₁ is moving with a velocity v₁ and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum but their different kinetic energies are E₁ and E₂ respectively. If m₁ > m₂ then:

⚔️ Practice NEET Physics free + battle 1v1 →