Question

A bead with mass 2.00 × 10-2 kg is moving along a wire in the positive direction of an x axis. Beginning at time t = 0, when the bead passes through x = 0 with speed 12.0 m/s, a constant force acts on the bead. The figure indicates the bead's position at times t0 = 0.00 s, t1 = 1.00 s, t2 = 2.00 s, and t3 = 3.00 s. The bead momentarily stops at t = 3.00 s. What is the kinetic energy of the bead at t = 10.0 s?

Answer 1

Answer on Question # 84070, Physics / Molecular Physics | Thermodynamics

Question 1. A bead with mass $2\cdot 10^{-2}\,kg$ is moving along a wire in the positive direction of an $x$ axis. Beginning at time $t=0$, when the bead passes through $x=0$ with speed $12.0\,m/s$, a constant force acts on the bead. The figure indicates the bead’s position at times $t_{0}=0\,s,\;t_{1}=1\,s,\;t_{2}=2\,s$, and $t_{3}=3\,s$. The bead momentarily stops at $t=3\,s$. What is the kinetic energy of the bead at $t=10\,s$?

Solution. Consider the point $t=3\,s$: the acceleration must be $a=\frac{\Delta v}{\Delta t}=\frac{-12}{3}=-4\,m/s^{2}$. So the particle begins at rest at $t=3\,s$.

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

$E_{k}=\frac{mv^{2}}{2}=\frac{0.02\cdot(-28)^{2}}{2}=7.84\,J.$

$\square$