A car starts from rest and accelerates to 90 km per h in 20 s. Find the force exerted by
the car and the work done.
Answers
Answer:
We have discussed how to find the work done on a particle by the forces that act on it, but how is that work manifested in the motion of the particle? According to Newton’s second law of motion, the sum of all the forces acting on a particle, or the net force, determines the rate of change in the momentum of the particle, or its motion. Therefore, we should consider the work done by all the forces acting on a particle, or the net work, to see what effect it has on the particle’s motion.
Let’s start by looking at the net work done on a particle as it moves over an infinitesimal displacement, which is the dot product of the net force and the displacement:
\[d{W}_{\text{net}}={\overset{\to }{F}}_{\text{net}}·d\overset{\to }{r}.\]
Newton’s second law tells us that
\[{\overset{\to }{F}}_{\text{net}}=m(d\overset{\to }{v}\text{/}dt),\]
, so
\[d{W}_{\text{net}}=m(d\overset{\to }{v}\text{/}dt)·d\overset{\to }{r}.\]
For the mathematical functions describing the motion of a physical particle, we can rearrange the differentials dt, etc., as algebraic quantities in this expression, that is,
\[d{W}_{\text{net}}=m(\frac{d\overset{\to }{v}}{dt})·d\overset{\to }{r}=md\overset{\to }{v}·(\frac{d\overset{\to }{r}}{dt})=m\overset{\to }{v}·d\overset{\to }{v},\]
where we substituted the velocity for the time derivative of the displacement and used the commutative property of the dot product [(Equation 2.30)]. Since derivatives and integrals of scalars are probably more familiar to you at this point, we express the dot product in terms of Cartesian coordinates before we integrate between any two points A and B on the particle’s trajectory. This gives us the net work done on the particle:
\[\begin{array}{cc}\hfill {W}_{\text{net},AB}& ={\int }_{A}^{B}(m{v}_{x}d{v}_{x}+m{v}_{y}d{v}_{y}+m{v}_{z}d{v}_{z})\hfill \\ & =\frac{1}{2}m{|{v}_{x}^{2}+{v}_{y}^{2}+{v}_{z}^{2}|}_{A}^{B}={|\frac{1}{2}m{v}^{2}|}_{A}^{B}={K}_{B}-{K}_{A}.\hfill \end{array}\]