Question

“The WE theorem may be viewed as a scalar form of the second Law.” Explain.

Answer 1

Hint Here first we find the force using the second law of motion and then put the value of this force in the definition of work done. From there using a little bit of mathematics and rearranging we can deduce the relation.

Formula Used: The formulae used in solving this question are
 $\Rightarrow \vec F = \dfrac{{d\vec p}}{{dt}} $ here, $ \vec F $ is the force acting on the particle, $ \vec p $ is the momentum of the particle, and $ \dfrac{d}{{dt}} $ is the differential operator.
 $\Rightarrow W = \int {\vec F.d\vec r} $ Here, $ W $ is the work done by the particle against the applied force, $ \vec F $ is the force applied on the particle, and $ \int {d\vec r} $ is the integration with respect to $ \vec r $ .

Complete step by step answer
We know that, according to Newton’s Second law of motion, a the momentum $ \vec p $ and the force $ \vec F $ are related by
 $\Rightarrow \vec F = \dfrac{{d\vec p}}{{dt}} $
We also know that momentum is given as,
 $\Rightarrow \vec p = m\vec v $
Where, $ \vec p $ is the momentum of the particle, $ m $ is the mass of the particle, and $ \vec v $ is the velocity of the particle.
Note that here $ \vec F,\vec p $ and $ \vec v $ are vector quantities.
Now, putting the value of $ \vec p $ in $ \vec F $ , we get
 $\Rightarrow \vec F = \dfrac{d}{{dt}}(m\vec v) $
If mass is independent of time, then
 $\Rightarrow \vec F = m\dfrac{{d\vec v}}{{dt}} $
Now, we know that work done by the particle in moving from point A to B is defined as,
 $\Rightarrow W = \int\limits_A^B {\vec F.d\vec r} $
Here, remember that work done is a scalar quantity. So, now using the value of $ \vec F $ found above, we can write,
 $\Rightarrow W = \int\limits_A^B {(m\dfrac{{d\vec v}}{{dt}}).d\vec r} \\
   \Rightarrow W = m\int\limits_A^B {(\dfrac{{d\vec v}}{{dt}}).d\vec r} \\
 $
This can be further re-written as,
 $\Rightarrow W = m\int\limits_A^B {(\dfrac{{d\vec r}}{{dt}}).d\vec v} $
Since, $ \vec v = \dfrac{{d\vec r}}{{dt}} $ so we can write,
 $\Rightarrow W = m\int\limits_A^B {\vec v \cdot d\vec v} $
Solving the above integral, we get
 $\Rightarrow W = \dfrac{1}{2}mv_B^2 - \dfrac{1}{2}mv_A^2 $
This expression obtained is the work energy theorem.
Thus, the second law of motion, which relates the vector quantities, can be easily simplified into W-E Theorem, in which each quantity is a scalar quantity.
 $ \therefore $ We can say that the W-E theorem is a scalar form of Newton’s 2nd Law of motion.

Note
The work energy theorem can be very easily derived using the third kinematic equation of motion. The above derivation is only used to show the relation of this theorem with Newton's second law of motion.