Problem — True or false on work and kinetic energy

Problem

True or false?

• (a) Work can be negative.
• (b) Kinetic energy can be negative.
• (c) A force perpendicular to the displacement does zero work.
• (d) If kinetic energy doubles, so does the speed.

Solution

Recall: $L = F\,s\,\cos\theta$ and $E_k = \tfrac{1}{2}mv^2$.

(a) TRUE. When the force has a component opposite to the displacement ($90^\circ < \theta \leq 180^\circ$), $\cos\theta < 0$ and the work is negative. This is the typical case of friction, or of a braking force: they subtract kinetic energy from the body.

(b) FALSE. $E_k = \tfrac{1}{2}mv^2$ with $m > 0$ and $v^2 \geq 0$: kinetic energy is always positive or zero, never negative.

(c) TRUE. If $\theta = 90^\circ$ then $\cos 90^\circ = 0$, so $L = 0$. A force perpendicular to the displacement (for example, the centripetal force) does no work.

(d) FALSE. Kinetic energy depends on the square of the speed: $E_k \propto v^2$. If $E_k$ doubles, then $v^2$ doubles and the speed becomes $v\sqrt{2} \approx 1{,}41\,v$, i.e. it increases by only about $41\%$. To double the speed you would need to quadruple the kinetic energy.

$\ev{\text{(a) T}\quad\text{(b) F}\quad\text{(c) T}\quad\text{(d) F}}$

Collegamenti

Argomenti: Lavoro ed energia Concetti: Lavoro di una forza · Energia cinetica Competenze: Ragionamento di ranking