Question

A 6 kg mass going 4 m/s is slowed to 3 m/s by a 2 N force for how much distance did the force act?

Answer 1

Given data:

* The mass of the object is m = 6 kg.

* The initial velocity of the object is u = 4m/s.

* The final velocity of the object is v = 3 m/s.

* The force acting on the object is F = 2 N.

Solution:

According to Newton's second law, the force acting on the object in terms of the acceleration of the object is,

`F=ma`

Substituting the known values,

`\begin{gathered} 2=6* a \\ a=(2)/(6) \\ a=0.33ms^(-2) \end{gathered}`

As the acceleration is decreasing the velocity, thus, the value of acceleration is,

`a=-0.33ms^(-2)`

By the kinematics equation, the distance traveled by the object in terms of the acceleration, initial velocity, and final velocity is,

`v^2-u^2=2aS`

where S is the distance traveled by the object,

Substituting the known values,

`\begin{gathered} 3^2-4^2=2*(-0.33)* S \\ 9-16=-0.66* S \\ -7=-0.66* S \\ S=(-7)/(-0.66) \end{gathered}`

By simplifying,

`\begin{gathered} S=(7)/(0.66) \\ S=10.6\text{ m} \end{gathered}`

Thus, the distance for which force act to change the velocity from 4 m/s to 3 m/s is 10.6 meters.