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

Options

A. `5 m/s`
B. `14 m/s`
C. `17.5 m/s`
D. `10 m/s`

Answer

`(4)`

Explanation

`U = (-7x + 24 y)J
vec F = -(dU)/(dx)hat i - (dU)/(dy) hat j = 7 hat i - 24 hat j`
` a_(x) = (F_(x))/(m) = 7/5 = 1.4 ms^(-2)`
Along positive x -axis
`a_(y) = (F_(y))/(m) = -24 /5 = - 4.8 ms^(-2)`
Along negative y -axis
`v_(x) = a_(x)t = 1.4 xx 2 = 2.8 ms^(-1)`
`v_(y) = a_(y) t = 4.8 xx 2 = 9.6 ms^(-1)`
`v = sqrt(v_(x)^(2) + v_(y)^(2)) = sqrt(2.8^(2) + 9.6 ^(2)) = 10 ms^(-1)`