Question 7.7: A 3.0 kg body is at rest on a frictionless horizontal air tr......
A 3.0 kg body is at rest on a frictionless horizontal air track when a constant horizontal force \vec{F} acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in Fig. 7- 25.The force is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F between t = 0 and t = 2.0 s?
Since this involves constant-acceleration motion, we can apply the equations of Table 2-1, such as x=\nu_0 t+\frac{1}{2}a t^2 (where x_0 = 0 ). We choose to analyze the third and fifth points, obtaining
\begin{aligned}& 0.2 m=\nu_0(1.0 s )+\frac{1}{2}a(1.0 s )^2 \\& 0.8 m=\nu_0(2.0 s )+\frac{1}{2}a(2.0 s )^2\end{aligned}.
Simultaneous solution of the equations leads to \nu_0=0 and a = 0.40m s² . We now have two ways to finish the problem. One is to compute force from F = ma and then obtain the work from Eq. 7-7. The other is to find ΔK as a way of computing W (in accordance with Eq. 7-10). In this latter approach, we find the velocity at t = 2.0 s from \nu=\nu_0+a t (so v = 0.80 m s) . Thus,
W=F d \cos \phi (work done by a constant force). (7-7)
\Delta K=K_f-K_i=W (7-10)
W=\Delta K=\frac{1}{2}(3.0 \,kg )(0.80 \,m / s )^2=0.96 \,J.
| Table 2-1 Equations for Motion with Constant Acceleration^{a} | ||
| 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 | ||