Power — Question 1

Question 1

Instantaneous power uses the velocity component along the force

At one instant a force on a particle is $(6,-2)$ N and its velocity is $(3,4)$ m/s. Later the velocity is $(-3,4)$ m/s while that force is unchanged.

1. Find the power delivered by this force at the first instant.

2. Find its power at the later instant.

3. Can either result by itself determine the rate of change of total kinetic energy?

Question 1 - Solution

Instantaneous power uses the velocity component along the force

(a) Take the local dot product.

The governing relation is $$\boxed{P=\frac{dW}{dt}=\vec F\cdot\vec v=F_xv_x+F_yv_y}$$

$$P=\vec F\cdot\vec v=6(3)-2(4)=10\ \mathrm W.$$ Multiplying the magnitudes would ignore the angle.

(b) Update the velocity, including signs.

$$P=6(-3)-2(4)=-26\ \mathrm W.$$ The force now removes kinetic energy through its own contribution.

(c) Sum over all forces.

$$dK/dt=\sum_j\vec F_j\cdot\vec v.$$ The computed result equals that rate only if all other powers sum to zero. A single positive contribution need not mean the particle speeds up.