Start Free Mock LoginAlready registered? Login here

Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Calculation of workEasy2 minQB

PhysicsEasysingle choice

A body of mass $0.5 kg$ travels on straight line path with velocity $v = (3 x^{2} + 4) m s^{- 1}$ . The net work done by the force during its displacement from $x = 0$ to $x = 2 m$ is

Options:

Answer:

Solution:

Given: $v = 3 x^{2} + 4$
Initial and final velocities are $v_{i} = 3 {(0)}^{2} + 4 = 4 m s^{- 1}$
and $v_{f} = 3 {(2)}^{2} + 4 = 16 m s^{- 1}$
From work-energy theorem,
$\Rightarrow Δ K E = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$\Rightarrow Δ K E = \frac{1}{2} \times \frac{1}{2} (16^{2} - 4^{2})$
$\Rightarrow Δ K E = W = 60 J$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Calculation of work

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...