Question

A force of 4N acts on a body at rest of 2kg for 4sec .after next 4sec 2N force operates for 2 sec. Calculate the distance travelled in these 10 sec​

Answer 1

Answer:

A force of 4N acts on a body of mass 2kg for 4 seconds.

mass of body, m = 2kg

initial velocity of body , u = 0 m/s

force acts on the body , F = 4N

from Newton's 2nd law,

F = ma

4 = 2a => a = 2m/s²

force acting on body for 4sec.

so, velocity of body changes in 4 sec.

use formula, v = u + at

v = 0 + 2 × 4 = 8 m/s

hence, body moves with velocity 8m/s . now, applied force is removed after 4sec then, body moves with constant velocity. hence, acceleration = 0

we have to find distance covered in 10 sec after force starts acting on it.

so, total distance = distance covered in first 4 sec + distance covered in rest 6 sec

= {0 × 4 + 1/2 × 2 × 4²} + {8 × 6 + 1/2 × 0 × 6²}

= 16 + 48 = 64m

Answer 2

Answer:

Step-by-step explanation:

QGP Maths AryaBhatta

Force = F = 4 N

Mass = m = 2 kg

Time = t = 4 sec

Initial velocity = u = 0

Let final velocity be v

A constant force acts on the body. So, the body has uniform accelerating.

Acceleration a = F/m = 4/2 = 2 m/s^2

Now, v = u + at

So, v = 0 + (2)(4)

So, v = 8 m/s

Thus, final velocity of the body is 8 m/s