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

PhysicsW.E.P.Work kinetic energy theoremEasy2 minPYQ_2020

PhysicsEasynumerical

An object of mass $8 kg$ starts to move from rest under the action of a variable force $F = 3 x N$ where $x$ is the distance (in $m$ ) covered by the object. If initially, the position of the object is $x = 2 m$ , then find its speed (in $m s^{- 1}$ ) when it crosses $x = 10 m$ .

Answer:

6.00

Solution:

Increment in kinetic energy = work done

$\frac{1}{2} m (v^{2} - u^{2}) = \int_{x_{1}}^{x_{2}} F . d x = \int_{2}^{10} (3 x) d x$

$\frac{1}{2} m v^{2} = \frac{3}{2} {[x^{2}]}_{2}^{10} = \frac{3}{2} [100 - 4]$

$\frac{1}{2} \times 8 \times v^{2} = \frac{3}{2} \times 96$

$v = 6 m s^{- 1}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Work kinetic energy theorem

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...