A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
Given: The mass of the body is 0.40 kg , the initial speed of the body is 10 m / s towards the north and the force acting on the body towards the south for 30 s is 8.0 N , the position of the body at time t = 0 is x = 0 and the force is applied from an instant t = 0 .
As the force is applied at the instant t = 0 therefore, the acceleration of the body at = − 5 s is zero.
The position of the body at t = − 5 s can be calculated using second equation of motion.
s 1 = u t + 1 2 a t 2 (1)
Where, u is the initial speed and a is the acceleration.
By substituting the values in the above equation, we get
s 1 = 10 ( − 5 ) + 1 2 ( 0 ) ( − 5 ) 2 = − 50 m
Thus, the position of the object at t = − 5 s is − 50 m .
From Newton’s second law, the acceleration of the object after t = 0 s is given as,
a = F m
Where, F is the force applied and m is the mass of the object.
By substituting the values in the above expression, we get
a = − 8.0 0.40 = − 20 m/s 2
By substituting this value in equation (1) at t = 25 s , we get the position of the body as,
s 2 = 10 ( 25 ) + 1 2 ( − 20 ) ( 25 ) 2 = − 6000 m = − 6 km
Thus, the position of the object at t = 25 s is − 6 km .
As the force is applied for 30 s therefore, the acceleration works only for this time period.
The position of the body at t = 30 s can be calculated from equation (1).
s 3 = 10 × 30 + 1 2 ( − 20 ) ( 30 ) 2 = − 8700 m
The velocity of the body at t = 30 s can be calculated as,
v = u + a t
By substituting the values in the above equation, we get
v = 10 + ( − 20 ) ( 30 ) = − 590 m/s
For the next 100 − 30 = 70 s , body moves with this uniform velocity.
The position of the body at t = 100 s can be calculated as,
s 4 = s 3 + v t ′
By substituting the values in the above equation, we get
s 4 = − 8700 + ( − 590 ) ( 70 ) = − 50000 m = − 50 km
Thus, the position of the object at t = 100 s is − 50 km .
Given: The mass of the body is
As the force is applied at the instant
The position of the body at
Where,
By substituting the values in the above equation, we get
Thus, the position of the object at
From Newton’s second law, the acceleration of the object after
Where,
By substituting the values in the above expression, we get
By substituting this value in equation (1) at
Thus, the position of the object at
As the force is applied for
The position of the body at
The velocity of the body at
By substituting the values in the above equation, we get
For the next
The position of the body at
By substituting the values in the above equation, we get
Thus, the position of the object at
