Question

a constant retarting force of 100N is applying to a body of mass 10kg moving inertially with speed of 10ms- how long the body take to stop​

Answer 1

Answer:

Explanation:This problem requires a simple application of Newton’s 2nd Law: F=ma, and a knowledge of 1D mechanics.

Let’s assume that the body is moving 15msin the positive x direction. The force is -50 N. This means the force has a magnitude of 50 newtons in the negative x direction.

Remember that 1N=1kg⋅ms2 , or one newton is equivalent to one kilogram-meter per second squared. This is useful because no conversions are required.

Let’s plug in our known values into Newton’s 2nd Law to find the acceleration:

−50N=20kg⋅a

−50kg⋅ms2=20kg⋅a

a=−2.5ms2

This means that the retarding force decelerates the body at −2.5ms2

Acceleration is the change in velocity. Velocity in terms of acceleration is:

vx=v0x+axt

Where v0x is velocity at time 0 (initial velocity) and t is the time passed. All of these have x’s in them because all of our motion is in the x direction (when motion happens in one dimension, we default to x and not y or z)

The body stops when vx=0

0ms=15ms+(t seconds)⋅(−2.5ms2)

−15ms=(t seconds)⋅(−2.5ms2)

t=6s

The body stops after 6 seconds.

Hope this helped!

Answer 2

Answer:

1 second

Explanation:

F =100 N (retardation)

m = 10 kg

since F = ma

so, a =F/m = 100/10 = 10 (deceleration)

As u = 10 m/s

v = 0m/s at it stops

a = -10 m/s^2

and v = u + at

so, t = v-u/a = -10/-10 = 1 sec