Question

A particle of mass 1kg is acted upon by a force 'F' which varies as shown in the figure. If initial
velocity of the particle is 10 m/s, the maximum velocity attained by the particle during the period is
F
20N
110
#
(0,0)
20
t(sec)
10N​

Answer 1

A particle of mass, m = 1kg is acted upon by a force F which varies as shown in the figure. initial velocity of particle, u = 10 m/s.

To find : The maximum velocity attained by the particle during the period.

solution : from figure, it is clear that, force is directly proportional to time ( 0 ≤ t ≤ 10sec). i.e., acceleration is also proportional to time (means, acceleration is increasing with time). it shows velocity of particle is increasing during 0 to 10s.

so particle attained maximum velocity at t = 10s

relation between F and t , F = (20N-0)/(10s - 0) t

⇒F = 2t

⇒ma = 1kg × a = 2t

⇒a = 2t m/s²

as we know, acceleration is the rate of change of velocity with respect to time.

so, a = dv/dt = 2t

⇒∫dv = 2∫t dt

⇒v - u = $[t²]^10_0$

⇒v - 10 = 100

⇒v = 110 m/s

Therefore the maximum velocity attained by particle is 110 m/s.