The displacement of a body of mass 2kg varies with time as $S={{t}^{2}}+2t$ where S is in meters and $t$ is in seconds. The work done by all the forces acting on the body during the time interval t=2s to 4s is.
Answer
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Hint: As a first step, one could read the question well and hence note down the given points that seem important. Then, one could find the velocity from the given displacement. Then find the kinetic energy for the respective time mentioned. After that apply the work energy theorem and hence find the answer.
Formula used:
Velocity,
$v=\dfrac{dS}{dt}$
Work energy theorem,
$W=\Delta KE$
Complete answer:
In the question, we are given the displacement of a body of mass 2kg to be vary with time as $S={{t}^{2}}+2t$………………………………………… (1)
Here, displacement is measured in meters and time is measured in seconds.
We are supposed to find the work done by all the forces acting on the body during the interval of t=2s to t=4s.
We know that the time rate of change of displacement will give the velocity. So, differentiating (1) we get,
$v=\dfrac{dS}{dt}=2t+2$……………………………………. (2)
Kinetic energy at t=2s would be,
$KE=\dfrac{1}{2}m{{\left( v\left( t=2 \right) \right)}^{2}}=\dfrac{1}{2}\times 2\times {{\left( \left( 2\times 2 \right)+2 \right)}^{2}}$
$\Rightarrow KE=36J$………………………………… (3)
Kinetic energy at t=4s would be,
$KE'=\dfrac{1}{2}m{{\left( v'\left( t=4 \right) \right)}^{2}}=\dfrac{1}{2}\times 2\times {{\left( \left( 2\times 4 \right)+2 \right)}^{2}}$
$\Rightarrow KE'=100J$……………………………………. (4)
Now we know from the work energy theorem that work done by external force would be equal to the change in kinetic energy. That is,
$W=\Delta KE=KE'-KE$
$\Rightarrow W=100J-36J$
$\therefore W=64J$
Hence, we found the work done in the given case to be 64J.
Note: Always keep in mind that the work energy theorem is exclusively applicable to the net work done and hence clearly, it has nothing to do with the work done by a single force. The SI unit of net work done is given by joules or newton meter. Another name given to this theorem is ‘work energy principle’.
Formula used:
Velocity,
$v=\dfrac{dS}{dt}$
Work energy theorem,
$W=\Delta KE$
Complete answer:
In the question, we are given the displacement of a body of mass 2kg to be vary with time as $S={{t}^{2}}+2t$………………………………………… (1)
Here, displacement is measured in meters and time is measured in seconds.
We are supposed to find the work done by all the forces acting on the body during the interval of t=2s to t=4s.
We know that the time rate of change of displacement will give the velocity. So, differentiating (1) we get,
$v=\dfrac{dS}{dt}=2t+2$……………………………………. (2)
Kinetic energy at t=2s would be,
$KE=\dfrac{1}{2}m{{\left( v\left( t=2 \right) \right)}^{2}}=\dfrac{1}{2}\times 2\times {{\left( \left( 2\times 2 \right)+2 \right)}^{2}}$
$\Rightarrow KE=36J$………………………………… (3)
Kinetic energy at t=4s would be,
$KE'=\dfrac{1}{2}m{{\left( v'\left( t=4 \right) \right)}^{2}}=\dfrac{1}{2}\times 2\times {{\left( \left( 2\times 4 \right)+2 \right)}^{2}}$
$\Rightarrow KE'=100J$……………………………………. (4)
Now we know from the work energy theorem that work done by external force would be equal to the change in kinetic energy. That is,
$W=\Delta KE=KE'-KE$
$\Rightarrow W=100J-36J$
$\therefore W=64J$
Hence, we found the work done in the given case to be 64J.
Note: Always keep in mind that the work energy theorem is exclusively applicable to the net work done and hence clearly, it has nothing to do with the work done by a single force. The SI unit of net work done is given by joules or newton meter. Another name given to this theorem is ‘work energy principle’.
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