NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
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 . x and y are in meter. If the particle starts from rest from origin then speed of the particle at t = 2 s is
Options
- A5   m   s - 1
- B14   m   s - 1
- C17.5   m   s - 1
- D10   m   s - 1
Correct answer
D. 10   m   s - 1
Step-by-step solution
F =   - ∂ U ∂ x   i ^ - ∂ U ∂ y   j ^ = 7 i ^ - 24 j ^ ∴ a x = F x m = 7 5 = 1.4   m   s - 2 a y = F y m =   - 24 5 = 4 .8   m   s - 2 along negative y-axis ∴ v x = a x t = 1.4   × 2 = 2 .8   m   s - 1 along negative y-axis And v y = 4.8   × 2 = 9 .6   m   s - 1 ∴ v =   v x 2 + v y 2 = 10   m   s - 1