
A body of mass 1 kg is kept at rest. A constant force of 6.0 N starts acting on it. Find the time taken by the body to move through a distance of 12 m.
Answer
517.3k+ views
Hint: From the given mass and force, we can find out the acceleration of the given mass. Then we can make use of one of the equations of motion with the known values, solving the equation, we can get the time taken by the body to move through the given distance.
Formula used:
Formula used:
Newton’s second law of motion is given as
$F = ma$
Equations of motion for a mechanical system are given as follows:
$
v = u + at \\
S = ut + \dfrac{1}{2}a{t^2} \\
{v^2} - {u^2} = 2aS \\
$
Complete step-by-step answer:
We are given a body whose mass is given as
$m = 1kg$
This body is kept at rest so its initial velocity is given as
$u = 0m/s$
A constant force is acting on this body whose magnitude is given as
$F = 6N$
Now from Newton’s second law of motion we can find out the acceleration of the given mass. This can be done in the following way.
$
F = ma \\
a = \dfrac{F}{m} \\
$
Inserting the value of force and mass, we get
$a = \dfrac{6}{1} = 6m/{s^2}$
We want to move this body using the above force through a distance which is given as
$S = 12m$
So we can make use of the second equation of motion in order to obtain the time taken to move the mass through the above distance.
$S = ut + \dfrac{1}{2}a{t^2}$
Inserting the known values, we get
$
12 = 0 + \dfrac{1}{2} \times 6 \times {t^2} \\
{t^2} = \dfrac{{12}}{3} = 4 \\
t = 2s \\
$
This is the required value of the time. Therefore, it takes 2 seconds for the body to move through a distance of 12 m under the action of a constant force of 6N.
Note: We can also define the force acting on a given mass as the rate of change of the momentum of the given mass. A constant force acting on a mass means that the acceleration of that mass is constant or the change in momentum is taking place by a constant value.
Formula used:
Formula used:
Newton’s second law of motion is given as
$F = ma$
Equations of motion for a mechanical system are given as follows:
$
v = u + at \\
S = ut + \dfrac{1}{2}a{t^2} \\
{v^2} - {u^2} = 2aS \\
$
Complete step-by-step answer:
We are given a body whose mass is given as
$m = 1kg$
This body is kept at rest so its initial velocity is given as
$u = 0m/s$
A constant force is acting on this body whose magnitude is given as
$F = 6N$
Now from Newton’s second law of motion we can find out the acceleration of the given mass. This can be done in the following way.
$
F = ma \\
a = \dfrac{F}{m} \\
$
Inserting the value of force and mass, we get
$a = \dfrac{6}{1} = 6m/{s^2}$
We want to move this body using the above force through a distance which is given as
$S = 12m$
So we can make use of the second equation of motion in order to obtain the time taken to move the mass through the above distance.
$S = ut + \dfrac{1}{2}a{t^2}$
Inserting the known values, we get
$
12 = 0 + \dfrac{1}{2} \times 6 \times {t^2} \\
{t^2} = \dfrac{{12}}{3} = 4 \\
t = 2s \\
$
This is the required value of the time. Therefore, it takes 2 seconds for the body to move through a distance of 12 m under the action of a constant force of 6N.
Note: We can also define the force acting on a given mass as the rate of change of the momentum of the given mass. A constant force acting on a mass means that the acceleration of that mass is constant or the change in momentum is taking place by a constant value.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

