Question 7.6: A bead with mass 1.8 × 10^-2 kg is moving along a wire in th......
A bead with mass 1.8 × 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 m/s, a constant force acts on the bead. Figure 7-24 indicates the bead’s position at these four times: t_0 = 0, t_1 = 1.0 s, t_2 = 2.0 s, and t_3 = 3.0 s. The bead momentarily stops at t = 3.0 s. What is the kinetic energy of the bead at t = 10 s?
We apply the equation x(t)=x_0+\nu_0 t+\frac{1}{2}a t^2, found in Table 2-1. Since at t = 0 s, x_0 = 0, and \nu_0=12 m/s , the equation becomes (in unit of meters)
x(t)=12 t+\frac{1}{2}a t^2.
With x = 10 m when t = 1.0 s , the acceleration is found to be a = -4.0 m/s² . The fact that a < 0 implies that the bead is decelerating. Thus, the position is described by x(t) = 12t -2.0t² . Differentiating x with respect to t then yields
\nu(t)=\frac{d x}{d t}=12-4.0 t.
Indeed at t =3.0 s, v(t = 3.0) = 0and the bead stops momentarily. The speed at t = 10 s is v(t = 10) = -28 m/s , and the corresponding kinetic energy is
K=\frac{1}{2}m \nu^2=\frac{1}{2}\left(1.8 \times 10^{-2}\,kg \right)(-28 \,m / s )^2=7.1 \,J.
| Table 2-1 Equations for Motion with Constant Acceleration^{a} |
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| Equation Number | Equation | Missing Quantity |
| 2-11 | \nu=\nu_{0}+a t | x-x_{0} |
| 2-15 | x-x_{0}=\nu_{0}t+{\textstyle{\frac{1}{2}}}a t^{2} | \boldsymbol{\nu} |
| 2-16 | \nu^{2}=\nu_{0}^{2}+2a(x-x_{0}) | t |
| 2-17 | x-x_{0}={\textstyle{\frac{1}{2}}}(\nu_{0}+\nu)t | a |
| 2-18 | x-x_{0}=\,\nu t-{\textstyle\frac{1}{2}}a t^{2} | \nu_{0}\, |
| ^{a}Make sure that the acceleration is indeed constant before using the equations in this table | ||