The velocity of a particle moving along the x-axis changes from $\text V_i$ to $\text V_f$. For which values of $\text V_i$ and $\text V_f$ is the total work done on the particle negative?
$$A) \; \text V_i = 2 \text{m/s}, \; \text V_f = 5\text{m/s} \quad C) \; \text V_i = -2 \text{m/s}, \text V_f = -5 \text{m/s}$$
$$B) \; \text V_i = -2 \text{m/s}, \; \text V_f = 5 \text{m/s} \quad D) \; \text V_i = -5 \text{m/s}, \; \text V_f = 2 \text{m/s}$$

Question

The velocity of a particle moving along the x-axis changes from $\text V_i$ to $\text V_f$. For which values of $\text V_i$ and $\text V_f$ is the total work done on the particle negative?
$$A) \; \text V_i = 2 \text{m/s}, \; \text V_f = 5\text{m/s} \quad C) \; \text V_i = -2 \text{m/s}, \text V_f = -5 \text{m/s}$$
$$B) \; \text V_i = -2 \text{m/s}, \; \text V_f = 5 \text{m/s} \quad D) \; \text V_i = -5 \text{m/s}, \; \text V_f = 2 \text{m/s}$$

lgbasallote · Answer

*why* isn't it C?

anonymous · Answer

it should be c :S

lgbasallote · Answer

sadly, you and i are both wrong

anonymous · Answer

wait a moment..
$$\Delta KE = Work done$$

lgbasallote · Answer

...that means what..?

lgbasallote · Answer

$$\frac 12 mv_f ^2 - \frac 12 mv_i ^2$$
right?

anonymous · Answer

$$\frac{ 1 }{ 2 } m (V_f^2-V_i^2)$$

lgbasallote · Answer

yes...that's what i said

anonymous · Answer

so if u square both.. work done is positive..

anonymous · Answer

The answer is D because in it the particle SLOWS down

anonymous · Answer

it mist be D

anonymous · Answer

2^2-5^2 = negative

lgbasallote · Answer

so it's D becaise the $\textbf{*magnitude*}$ of the final velocity is smaller than the magnitude of the initial velocity?

anonymous · Answer

Work is quite intuitive it EnergyFinal - Energy Initial

anonymous · Answer

Yes this depends only on MAGNITUDE

lgbasallote · Answer

oh. would've been fun to have known that earlier...

anonymous · Answer

M v^2 only depends on magnitude

anonymous · Answer

Mdl ?