Question

11. A force of 2 N acts for 5 seconds on a particle of mass 0.5 kg initially at rest. Calculate the distance moved by the particle in these five seconds and (ii) next 5 seconds (3) 12. A 8000 kg engine pulls a train of 5 wagons, cach of 2000 kg along a horizontal track. If the

Answer 1

♡ Heya user ♡

Given :-

✰ F = 2N, m=0.5kg, u=0 & t=5sec

Solution :-

(i) At first, we need to find the value of( a) acceleration.

Hence, a = F/m = 2/0.5 = 4m/s^2

As we know, s = ut + 1/2 at^2

= 0+1/2 *4*(5)^2

=50 m

(ii) According to the question,

Velocity acquired at the end of 5sec is, v

= u+at

= 0+4*5

=20 m/s

Hence, distance moved in next 5sec is

= v*t

= 20*5

= 100m

Here , remember that the velocity is uniformed.

I am sorry but ur 2nd question isn't a complete question.

✧Thanks ✧