Answer
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Hint: In simple words work done is a product of force and displacement, only if we remove the dot product. Force is defined as a product if mass and acceleration and acceleration is defined as velocity per unit time. You can relate this theorem to examples in your daily life.
Complete step by step by solution:
Statement: “Work done by a force in displacing a body measures the change in kinetic energy of the body.”
According to work energy theorem, when a force does some work on a body, the kinetic energy of the body increases by the same amount conversely when an opposing force is applied on a body. Its kinetic energy decreases. This decrease in kinetic energy of the bodies equal to the work done by the body against the opposing force. Thus work and kinetic energy are equivalent quantities. Here we have to assume the work done by the force is only effective in changing kinetic energy of the body but on potential energy of the body.
Work energy theorem
Consider a body of mass 'm' moving with initial velocity 'u' is acted upon by a force f in the direction of motion of the body.
Let vector ds be the displacement of the body in the direction of force and vector v the final velocity in time dt.
Small amount of work done by force is,
$\begin{align}
& dw=\overrightarrow{f}.\overrightarrow{ds}=fds \\
& Therefore,dw=fds=mada \\
& Therefore,dw=m\dfrac{dv}{dt}.ds=m(\dfrac{ds}{ds})dv \\
& dw=mvds \\
\end{align}$
Therefore dw = f ds = ma ds
Total work done by the force in increasing the velocity of the body form u to v is
\[w=\int\limits_{u}^{v}{dw}=\int\limits_{u}^{v}{mvdv=\dfrac{1}{2}m[{{v}^{2}}-{{u}^{2}}]}\]
w = final kinetic energy – initial kinetic energy
w = change in kinetic energy.
Note: Work done by the force is the measure of change in kinetic energy of the body which proves the work energy theorem. Integration shows the path of work done from initial position to final position.
Complete step by step by solution:
Statement: “Work done by a force in displacing a body measures the change in kinetic energy of the body.”
According to work energy theorem, when a force does some work on a body, the kinetic energy of the body increases by the same amount conversely when an opposing force is applied on a body. Its kinetic energy decreases. This decrease in kinetic energy of the bodies equal to the work done by the body against the opposing force. Thus work and kinetic energy are equivalent quantities. Here we have to assume the work done by the force is only effective in changing kinetic energy of the body but on potential energy of the body.
Work energy theorem
Consider a body of mass 'm' moving with initial velocity 'u' is acted upon by a force f in the direction of motion of the body.
Let vector ds be the displacement of the body in the direction of force and vector v the final velocity in time dt.
Small amount of work done by force is,
$\begin{align}
& dw=\overrightarrow{f}.\overrightarrow{ds}=fds \\
& Therefore,dw=fds=mada \\
& Therefore,dw=m\dfrac{dv}{dt}.ds=m(\dfrac{ds}{ds})dv \\
& dw=mvds \\
\end{align}$
Therefore dw = f ds = ma ds
Total work done by the force in increasing the velocity of the body form u to v is
\[w=\int\limits_{u}^{v}{dw}=\int\limits_{u}^{v}{mvdv=\dfrac{1}{2}m[{{v}^{2}}-{{u}^{2}}]}\]
w = final kinetic energy – initial kinetic energy
w = change in kinetic energy.
Note: Work done by the force is the measure of change in kinetic energy of the body which proves the work energy theorem. Integration shows the path of work done from initial position to final position.
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