Question

Kinetic energy of a particle moving along a circle of radius R depends on the distance covered as T = ks^2 where k is a constant. The Force acting on the particle as a function of s is..??

Answer 1

First workout an expression for the instantaneous speed:

Kinetic Energy = 1/2 m v *v
ks^2 = 1/2 m v ^2
=> v = (2k/m)^1/2 (s)


There are two components to the total Force :
1) Centripetal Force producing the circular motion .
=> CF = mv^2/R
= 2ks^2/R
= 2ks^2/R

2) Tangential Force :
There is tangential acceleration as kinetic energy is increasing as the particle moves .
Tangential motion:

$a \: = \frac{dv}{dt} = d( \sqrt{ \frac{2k}{m}} s) \div dt \\ \: \: \: \: \: \: \: \ = \sqrt{ \frac{2k}{m} } \times v \\ \: \: \: \: \: \: \: \: \: = \sqrt{ \frac{2k}{m} } \times \sqrt{ \frac{2k}{m} } \: s \\ \: \: \: \: \: \: \: \: \: = \frac{2k}{m} s$

Tangential Force = m a
=2ks


Now, Resultant Force =
$\sqrt{ {cf}^{2} + {tf}^{2} } \\ = \sqrt{ {(2ks)}^{2} + { \frac{(2k {s}^{2} )}{ {r}^{2} } }^{2} } \\ = 2ks \sqrt{1 + \frac{ {s}^{2} }{ {r}^{2} } } \\ where \: r \: is \: radius \: of \: circle.$

Hope, you understand my answer and it may helps you.

Answer 2

Answer:

Here is your answer for the question.

Explanation: