A force acts for 10 s on a stationary body of mass 100 kg after which the force cases to act. The body moves through a distance of 100 m in the next 5 s. Calculate:
(i) the velocity acquired by the body
(ii) the acceleration produced by the force, and
(iii) the magnitude of the force.
A force acts for 10 s on a stationary body of mass 100 kg after which the force cases to act. The body moves through a distance of 100 m in the next 5 s. Calculate:
(i) the velocity acquired by the body
(ii) the acceleration produced by the force, and
(iii) the magnitude of the force.
Given, m = 100 kg
Let F be the force that acts for t = 10 s
So, a = F/m ..........(1)
Using Ist eq. of motion :
v = u + at
v = 10 a ............(2) [ considering the body starts from rest]
After 10s, the force ceases and the body moves with velocity v. In 5 sec, it covers a distance of 100 m.
As, Distance = Speed X time
so, 100 = (10 a) X 5
or a = 2 m/s2 ........(3)
Putting this value of 'a' from eq. 3 into eq. 1, we get :
F = m a = 100 kg X 2 m/s2 = 200 N
Putting value of 'a' from eq. 3 into eq. 2, we get :
v = 10 a = 10 s X 2 m/s2 = 20 m/s
Given, m = 100 kg
Let F be the force that acts for t = 10 s
So, a = F/m ..........(1)
Using Ist eq. of motion :
v = u + at
v = 10 a ............(2) [ considering the body starts from rest]
After 10s, the force ceases and the body moves with velocity v. In 5 sec, it covers a distance of 100 m.
As, Distance = Speed X time
so, 100 = (10 a) X 5
or a = 2 m/s2 ........(3)
Putting this value of 'a' from eq. 3 into eq. 1, we get :
F = m a = 100 kg X 2 m/s2 = 200 N
Putting value of 'a' from eq. 3 into eq. 2, we get :
v = 10 a = 10 s X 2 m/s2 = 20 m/s
