Power — Question 7

Question 7

Average force times average speed can miss correlated variation

During the first second, an actuator exerts 2 N along a body moving at 1 m/s. During the second second, it exerts 6 N along the same body moving at 3 m/s. Other forces prescribe these speeds; ignore the transition interval.

1. Find the work of this actuator and its average power over two seconds.

2. Compare with the product of its time-averaged force and time-averaged speed.

3. Explain the discrepancy and give one condition that would make the product rule valid.

Question 7 - Solution

Average force times average speed can miss correlated variation

(a) Average the force-velocity product.

The governing relation is $$\boxed{\overline P=\frac1T\int_0^TF(t)v(t)\,dt}$$ Compute each stage’s power before averaging: $$P_1=2(1)=2\ \mathrm W,\qquad P_2=6(3)=18\ \mathrm W.$$ $$\boxed{W=2(1)+18(1)=20\ \mathrm J},\qquad\boxed{\overline P=20/2=10\ \mathrm W}.$$

(b) Compute the separate averages.

$$\overline F=(2+6)/2=4\ \mathrm N,\qquad \overline v=(1+3)/2=2\ \mathrm{m/s}.$$ Their product is 8 W, smaller than the actual average.

(c) Retain the correlation.

The difference is the force–speed covariance: $$\overline{Fv}-\overline F\,\overline v=10-8=2\ \mathrm W>0.$$ Larger force coincides with larger speed. A constant force or constant speed makes the covariance zero, restoring the product rule.