A vehicle of mass `m` starts moving along a horizontal circle of radius R such that its speed varies with distance s covered by the vehicle as `c=Ksqrts`, where K is a constant. Calculate:
a. Tangential and normal force on vehicle as fucntion of s.
b. Distance s in terms of time t.
c. Work done by the resultant force in first t seconds after the beginning of motion.

Answer

a. `(mK^2s)/(R)`; `1/2mK^2`; b. `1/4K^2t^2`; c. `1/8mK^4t^2`

Explanation

Since, the vehicle is moving along a circle, therefore, it is necessarily subjected to a centripetal acceleration.
The acceleration `a_n=(v^2)/(R)=(K^2s)/(R)`
The normal force `F_n=m.a_n=(mK^2s)/(R)`
Since, speed v of particle is increasing, therefore, it is necessary subjected to a tangerial acceleration also,
The acceleration `a_t=(d)/(dt)*v=(K)/(2sqrts)*(ds)/(dt)=(K)/(2sqrts)v=(K^2)/(2)`
Tangential forces, `F_t=m.a_t=1/2mK^2`
Since, s is the distance moved by the particle, therefore, speed v is `ds//dt`.
Hence, `(ds)/(dt)=Ksqrts`
`(ds)/(sqrts)=Kdt`
Integrating the above equation, `underset0oversets int(ds)/(sqrts)=Kint_0^tdt`
`2sqrts=Kt` or `s=1/4K^2t^2` `:. Speed v=Ksqrts=1/2K^2t`
Work done by resultant force is used to increase kinetic energy of the particle.
Kinetic energy at time t, `E=1/2mv^2=1/8mK^4t^2`
Initial kinetic energy, `E_0=0`.
Increase in kinetic energy
`=E-E_0=1/8mK^4t^2`