Question

Find the time for which a force of 1kgf
acts on a body of mass 1 kg moving
with a uniform speed of 4m/s to stop
the body. (given g=10 m/s2)​

Answer 1

Answer:

0.4 s

Step-by-step explanation:

Stopping force is F = – 10 N.

Here, F = m[(v - u)/t]

     t = 1 kg [(0 - 4 m/s)/-10 N]

       = 1 kg [0.4 m/s]

So, t = 0.4 sec.

Answer 2

Answer:

Time = 0.4s

Step-by-step explanation:

Stopping force is alternatively called Brake Power, the braking pressure refers to the whole quantity of pressure exerted on a transferring frame that reasons it to sluggish to a halt.

• The frame—or the vehicle—have to be journeying at a recognised consistent pace for this pressure to be calculated correctly.
• The braking force Fb is directly proportional to the speed of the vehicle and
• can be expressed by the relationship: Fb = kv.

Stopping force = -10N (Given)

Here , f = m a

where

$a= (v-u)/t$

hence ,

$f = m \times (v-u)/t$

Given ,

1. m = 1 kg
2. u= 4ms^-1
3. f= -10N

putting all these in the above equation:

t=

$1 \times \frac{(0 - 4)}{ - 10}$

$1 \times 0.4 \\ = 0.4$

hence ,

t= 0.4

(#SPJ2)