Answer to Question #181080 in Mechanics | Relativity for Geoffrey Genun

Question

positive direction of an x axis. Beginning at time t ! 0, when

the bead passes through x ! 0 with speed 12 m/s, a constant force

acts on the bead. Figure 7-24 indicates the bead’s position at

these four times: t0 ! 0, t1 ! 1.0 s, t2 ! 2.0 s, and t3 ! 3.0 s. The

bead momentarily stops at t ! 3.0 s.What is the kinetic energy of

the bead at t ! 10 s?

Answer 1

Consider the point t = 3 s: the acceleration must be $a= \frac{\triangle v}{\triangle t}=\frac{-12}{3}=-4 m/s^2$

So the particle begins at rest at t = 3 s

7 s later at t = 10 s we have velocity v = at = $-4 \times7 = -28 m/s.$

Assuming the bead had a mass of $2 \times 10^{-2}$ kg moving along a wire in the positive direction

of an x axis

$E_k= \frac{mv^2}{2}= \frac{0.02 \times (-28)^2}{2}=7.84 J$