Question

A particle of mass 0.01kg travels along a space curve with velocity given by 4i^+16k^ m/s. After some time it's velocity becomes 8i^+20k^ m/s due to the action of a conservative force. Find the work done on the particle during this interval of time?

Answer 1

i think you need using the Newton's second law ...

Answer 2

@DARno is off line

Answer 3

The easiest way is using the work-energy theorem, which reads:
\(W = \Delta K,\)
where \(W\) is the work done by the force and \(\Delta K\) is the change in the kinetic energy. Can you see how you can apply this?

I'll just leave the proof here, which is quite simple:
We start by writing that the work done by a force \(\vec{F}\) along and element of length \(\textrm{d}\vec{r}\) is:
\(\textrm{d}W = \vec{F}\cdot\textrm{d}\vec{r} = \vec{F} \cdot \vec{v} \textrm{ d}t,\)
where \(\vec{v} = \dfrac{\textrm{d}\vec{r}}{\textrm{d}t}\) is the velocity.

We now apply Newton's 2nd Law:
\(\vec{F} = m\dfrac{\textrm{d}\vec{v}}{\textrm{d}t}\)
and notice that:
\(\vec{F} \cdot \vec{v} = m \dfrac{\textrm{d}\vec{v}}{\textrm{d}t} \cdot \vec{v} = \dfrac{m}{2}\dfrac{\textrm{d}}{\textrm{d}t}(\vec{v}\cdot\vec{v}) = \dfrac{\textrm{d}}{\textrm{d}t}\left(\dfrac{1}{2}mv^2\right).\)

We now integrate both sides of the first equation for \(t \in [t_1,t_2]\):
\(\displaystyle W = \int \textrm{d}W = \int\limits_{t_1}^{t_2} \vec{F} \cdot \vec{v} \textrm{ d}t = \int\limits_{t_1}^{t_2} \dfrac{\textrm{d}}{\textrm{d}t}\left(\dfrac{1}{2}mv^2\right)\textrm{ d}t = \dfrac{1}{2}mv^2\Big\vert_{t_1}^{t_2} = \Delta K,\)
where \(\Delta K = \dfrac{1}{2}mv_2^2 - \dfrac{1}{2}mv_1^2 = K_2 - K_1\) is the change inthe kinetic energy \(K_i = \dfrac{1}{2}mv_i^2\).