Start Free Mock LoginAlready registered? Login here

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

PhysicsW.E.P.Calculation of workEasy2 minQB

PhysicsEasynumerical

A force of $F = (5 y + 20) \hat{j} N$ acts on a particle. The work done by this force when the particle is moved from $y = 0 m$ to $y = 10 m$ is ________ $J$ .

Answer:

450.00

Solution:

$F = (5 y + 20) \hat{j}$

$ω = \int F d y = \int_{0}^{10} (5 y + 20) dy$

$= {(\frac{5 y^{2}}{2} + 20 y)}_{0}^{10}$

$= \frac{5}{2} \times 100 + 20 \times 10$

$= 250 + 200 = 450 J$

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

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...