Question

A particle of mass 5 kg moves in a circle of radius 20 cm. Its linear speed at a time t is given by v = 4t, t is in the second and v is in ms^-1. Find the net force acting on the particle at t = 0.5 s.

Options

• 20 N

• `20sqrt23` N

• `20sqrt(26)` N

• 10 N

Answer 1

`underlinebb(20sqrt(26)  N)`

Explanation:

Given, r = 20 cm = 0.2 m, t = 0.5 s, v = 4t and m = 5 kg

Radial acceleration, `a_r = v^2/r = (4t)^2/0.2 = (16t^2)/0.2 = 80t^2`

= `80 xx (0.5)^2` = 20 ms^-2

Tangential acceleration of the particle,

`a_t = (dv)/dt = d/dt(4t)` = 4 ms^-2

∴ Net acceleration,

`a_n = sqrt(a_r^2 + a_t^2) = sqrt((20)^2 + (4)^2) = 4sqrt26` ms^-2

So, net force, `F_n = ma_n = 5 xx 4sqrt26 = 20sqrt26` N