A paticle of mass 50 g moves on a straight line. The vatiation of speed with time is shown in figure. Find the force acting on the particle at t=2,4 and 6 secons.

Explanation

Given `m= 50 g = 5xx 10^-2 kg`
As shown in the ure
Slopd of `OA = tan theta`
` =(AD)/(OD) = 15/3 = m/s^2`

So, at t=2 sec, acceleration is 5 m/s^2
Force `= ma= 5xx 10^-2 xx 5`
` = 0.25 N ` along the motion ltBrgt At `t=4 sec`,
Slope of `ab=0`
and acceleration =0 [Since `tan theta = 0^0]`
`t t=6 sec`,
acceleration = slope of BC
`In /_\ BEC, tan theta =5`
`Slope of BC = tan (108^0-theta)-tan theta`
` = 5 m/s^2 (deceleration )
`:. Force = ma= 5xx 10^-2 xx5` ltBrgt `=0.25 N`
Opposite to the motion.