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Home » A body of mass 0.40 kg moving initially with a constant speed of 10 ms-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.
A body of mass 0.40 kg moving initially with a constant speed of 10 ms-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.
CBSE | Class 11 | Laws of Motion | NCERT | Physics
A body of mass 0.40 kg moving initially with a constant speed of 10 ms-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,

Body mass is 0.40 kg.

u = 10 m/s initial velocity

f = -8 N force (retarding force)

Using the formula S = ut + (12) at2,

(a) At time t = – 5 s, position

From t = 0 s, the force acts on the body.

As a result, when the time is – 5 s, the body’s acceleration is 0.

S1= (10)(-5) + (½) (0) (-5)2 = – 50 m

(b) At time t = 25 s, position

The force acting in the opposite direction causes the body to accelerate.

a = F/a = -8 /0.4 = -20 ms-2

S2= (10)(25) + (½) (-20) (25)2 = – 6000 m

2 = − 6000 m

(c) At time t = 100 s, position

The body will move under the force’s retardation for the first 30 seconds, then the speed will remain constant.

As a result, distance covered in 30 sec

S3= (10)(30) + (½)(-20)(302)

= 300 – 9000 = – 8700m

The speed after 30 sec is

v = u + at

v = 10 – (20 x 30) = 590 m/s

The distance covered in the next 70 sec is

S4 = – 590 x 70 + (½) (0) (70)2 = – 41300 m

Therefore the position after 100 sec = S3 + S4 = – 8700 – 41300 = – 50000m

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