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

Question

The potential energy of a particle of mass 5 kg moving in the x- y plane is given by, U=(- 7 x + 24 y) J where x and y being in metre. If the particle starts from rest from the origin, then the speed of the particle at t=2 s is : (A) 5 ms- 1 (B) 14 ms- 1 (C) 17 ms- 1 (D) 10 ms- 1

Tardigrade Admin · Accepted Answer

∵ overset arrow F=-[(Δ U/Δ x) hati + (Δ U/Δ y) hatj]=-[- 7 hati + 24 hatj] ∴ F=√72 + 2 42=25 N⇒ a=F/m=25/5=5 ms- 2 ∵ v=u+at =0+5 × 2=10 ms- 1

Q. The potential energy of a particle of mass $5 k g$ moving in the $x - y$ plane is given by, $U = (- 7 x + 24 y) J$
where $x$ and $y$ being in metre. If the particle starts from rest from the origin, then the speed of the particle at $t = 2 s$ is :

$5 m s^{- 1}$

$14 m s^{- 1}$

$17 m s^{- 1}$

$10 m s^{- 1}$

$∵ F \to = - [\frac{Δ U}{Δ x} \hat{i} + \frac{Δ U}{Δ y} \hat{j}] = - [- 7 \hat{i} + 24 \hat{j}]$
$∴ F = 7^{2} + 2 4^{2} = 25 N \Rightarrow a = F / m = 25/5 = 5 m s^{- 2}$
$∵ v = u + a t$
$= 0 + 5 \times 2 = 10 m s^{- 1}$

Q. The potential energy of a particle of mass 5kg moving in the x−y plane is given by, U=(−7x+24y)J where x and y being in metre. If the particle starts from rest from the origin, then the speed of the particle at t=2s is :

5ms−1

14ms−1

17ms−1

10ms−1