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Question

Physics Question on work, energy and power

The potential energy of a particle of mass 5kg5 kg moving in the xyx-y plane is given by U=(7x+24y)Jx&yU=(-7 x+24 y) J \cdot x \& y being in metre. If the particle starts from rest from origin, then speed of particle at t=2st=2 s is :

A

5 m/s

B

14 m/s

C

175 m/s

D

10 m/s

Answer

10 m/s

Explanation

Solution

F=ui^xuj^yF=\frac{-\partial u \hat{i}}{\partial x} \frac{-\partial u \hat{j}}{\partial y} =7i^24j^=7 \hat{i}-24 \hat{j} ax=Fxm=75=1.4m/s2\therefore a_{x}=\frac{F_{x}}{m}=\frac{7}{5}=1.4\, m / s^{2} along positive xx- axis ay=Fym=245a_{y}=\frac{F_{y}}{m}=-\frac{24}{5} =4.8m/s=4.8\, m / s along negative y-axis vx=axt=1.4×2=2.8m/s\therefore v_{x}=a_{x} t=1.4 \times 2=2.8 \,m / s and vy=4.8×2=9.6m/sv_{y}=4.8 \times 2=9.6 \,m / s v=vx2+vy2=10m/s\therefore v=\sqrt{v_{x}^{2}+v_{y}^{2}}=10 \,m / s