Question
The potential energy of a particle of mass $5 \mathrm{~kg}$ moving in the $x-y$ plane is given by $U=(-7 x+24 y)$ J. $x$ and $y$ being in meter. If the particle starts from rest from origin then speed of particle at $t=2 \mathrm{sec}$ is:(a) $5 \mathrm{~m} / \mathrm{s}$(b) $14 \mathrm{~m} / \mathrm{s}$(c) $17.5 \mathrm{~m} / \mathrm{s}$(d) $10 \mathrm{~m} / \mathrm{s}$
Step 1
This gives us: \[F_x = -\frac{\partial U}{\partial x} = 7 \, \text{N}\] \[F_y = -\frac{\partial U}{\partial y} = -24 \, \text{N}\] Show more…
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The potential energy of a particle of mass $5 \mathrm{~kg}$ moving in the $x y$-plane is given by $U=(-7 x+24 y) \mathrm{J}, x$ and $y$ being in metre. If the particle starts from rest from origin, then speed of particle at $t=2 \mathrm{~s}$ is (a) $5 \mathrm{~ms}^{-1}$ (b) $01 \mathrm{~ms}^{-1}$ (c) $17.5 \mathrm{~ms}^{-1}$ (d) $10 \mathrm{~ms}^{-1}$
Work, Energy and Power
Round 1
The potential energy of a particle of mass $5 \mathrm{~kg}$ moving in the $x-y$ plane is given by $U=-7 x+24 y$ joule, $x$ and $y$ being in metre. Initially at $t=0$ the particle is at the origin and moving with velocity $(2 \hat{i}+3 \hat{j}) \mathrm{m} / \mathrm{s}$. The magnitude of the force on the particle and the acceleration of the particle are (a) $35 \mathrm{~N}, 7 \mathrm{~m} / \mathrm{s}^{2}$ (b) $30 \mathrm{~N}, 6 \mathrm{~m} / \mathrm{s}^{2}$ (c) $25 \mathrm{~N}, 5 \mathrm{~m} / \mathrm{s}^{2}$ (d) None of these
The potential energy of a particle of mass 5 kg moving in the XY plane is given by the equation U = (-7x + 24y) J, where x and y are in meters. If the particle starts its motion from the origin, then: (i) The magnitude of the acceleration of the particle is 5 m/s^2. (ii) The speed of the particle at t = 5 s is 25 m/s. (iii) The particle moves on a parabolic path in the XY-plane. (iv) The particle moves on a straight line in the XY-plane. A. (i, ii, iv) B. (i, iii) C. (ii, iv) D. (i, iii, iv)
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