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

Answer 1

QUESTION:

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 next 5 seconds.

ANSWER:

• The distance travelled by the particle in the first five seconds = 50 m.

• The distance travelled by the particle in next 5 seconds = 100 m.

GIVEN:

• A force of 2 N acts for 5 seconds on a particle of mass 0.5 kg initially at rest

TO FIND:

• The distance travelled by the particle in the first five seconds and next 5 seconds.

EXPLANATION:

$\bigstar \ \boxed{\bold{\large{\green{a = \dfrac{F}{m}}}}}$

$\implies \tt F=2\ N$

$\implies \tt m = 0.5 \ kg$

$\sf \leadsto a = \dfrac{2}{0.5}$

$\sf \leadsto a = \dfrac{20}{5}$

$\sf \leadsto a = 4 \ m {s}^{ - 2}$

$\bigstar \ \boxed{\bold{\large{\red{s = ut +\dfrac{1}{2} at^2}}}}$

$\tt \implies u = 0$

$\tt \implies t = 5 \ s$

$\tt \implies a= 4 \ m {s}^{ - 2}$

$\sf \leadsto s = 0(5)+\dfrac{1}{2} 4(5)^2$

$\sf \leadsto s = 0+2(25)$

$\sf \leadsto s = 50 \ m$

$\bf After \ 5 \ s, force \ is \ not \ applied \ \therefore \ a = 0$

$\bigstar \ \boxed{\bold{\large{\blue{v = u+at}}}}$

$\tt \implies u = 0$

$\tt \implies a= 4 \ m {s}^{ - 2}$

$\tt \implies t = 5 \ s$

$\sf \leadsto v = 0+4(5)$

$\sf \leadsto v =20 \ m {s}^{ - 1}$

$\textbf{This is the initial speed for next 5 s}$

$\bigstar \ \boxed{\bold{\large{\pink{s = ut +\dfrac{1}{2} at^2}}}}$

$\tt \implies u =20 \ m {s}^{ - 1}$

$\tt \implies t = 5 \ s$

$\tt \implies a= 0$

$\sf \leadsto s = 20(5) +\dfrac{1}{2} (0)(5)^2$

$\sf \leadsto s = 100 +0$

$\sf \leadsto s = 100 \ m$

Hence the distance travelled by the particle in the first five seconds = 50 m and the distance travelled in next 5 seconds = 100 m.

Answer 2

$\red{♡ Heya user♡}$

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,

(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.