Question 6.3: USING WORK AND ENERGY TO CALCULATE SPEED Let’s look again a...

IDENTIFY and SET UP:

We’ll use the work–energy theorem, Eq. (6.6), W_{\mathrm{tot}}=K_{2}-K_{1} since we are given the initial speed v_1 = 2.0 m/s and want to find the final speed v_2. Figure 6.11 shows our sketch of the situation. The motion is in the positive x-direction. In Example 6.2 we calculated the total work done by all the forces: W_{tot} = 10 kJ. Hence the kinetic energy of the sled and its load must increase by 10 kJ, and the speed of the sled must also increase.

EXECUTE:

To write expressions for the initial and final kinetic energies, we need the mass of the sled and load. The combined weight is 14,700 N, so the mass is

Then the initial kinetic energy K_1 is

K_{1}=\frac{1}{2} m v_{1}^{2}=\frac{1}{2}(1500 \mathrm{~kg})(2.0 \mathrm{~m} / \mathrm{s})^{2}=3000 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} = 3000 J

The final kinetic energy K_2 is

The work–energy theorem, Eq. (6.6), W_{\mathrm{tot}}=K_{2}-K_{1} gives

Setting these two expressions for K_2 equal, substituting 1 J = 1 kg . m^2/s^2, and solving for the final speed v_2, we find

EVALUATE: The total work is positive, so the kinetic energy increases (K_2> K_1) and the speed increases (v_2 > v_1).
This problem can also be solved without the work–energy theorem. We can find the acceleration from \sum \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\boldsymbol{a}} and then use the equations of motion for constant acceleration to find v_2.
Since the acceleration is along the x-axis, a=a_{x}=\frac{\sum F_{x}}{m}=\frac{500 \mathrm{~N}}{1500 \mathrm{~kg}}=0.333 \mathrm{~m} / \mathrm{s}^{2}

Then, using Eq. (2.13) (v_{x}^{2}=v_{0 x}^{2}+2 a_{x}\left(x-x_{0}\right)),

= 17.3 \mathrm{m}^{2} / \mathrm{s}^{2}

v_2 = 4.2 m/s

This is the same result we obtained with the work–energy approach, but there we avoided the intermediate step of finding the acceleration. You will find several other examples in this chapter and the next that can be done without using energy considerations but that are easier when energy methods are used. When a problem can be done by two methods, doing it by both methods (as we did here) is a good way to check your work.