Question

To practice Tactics Box 9.1 Calculating the Work Done by a Constant Force. Recall that the work W done by a constant force F⃗ at an angle θ to the displacement Δr⃗ is W=F⃗ ⋅ Δr⃗ =FΔrcosθ. The vector magnitudes F and Δr are always positive, so the sign of W is determined entirely by the angle θ between the force and the displacement.

Answer 1

Answer:

See espplanation below.

Explanation:

The deduction of the formula given is this one:

We can assume that we have a constant force F acting on a point that follows a curve Y with a velocity v at each instant, we can define an small amount of work $d W$ who occurs on a instant dt and we can calculate it as:

$d W= F dr = F v dt$

And we have that Fv represent the power at the instant dt and we can sum all the small amounts for the work in the entire trajectory with the following integral:

$W =\int_{t_1}^{t_2} F v dt = \int_{t_1}^{t_2} F \frac{dr}{dt} dt =\int_M F ds$

And M represent the trajectory from $y(t_1)$ to $y(t_2)$

If we assume that the force is constant so then we can do this with the integral:

$W= \int_M F dr = F \int_M dr = F \Delta r$

Where $\Delta r$ represent the displacement of the point along the curve.

We know that the fot product can be expressed on this way:

$F \Delta r = f Cos \theta dr$

Where $\theta$ represent the angle between the force vector and the direction of the movement, so in general we have this:

$W= \int_M F dr = F \Delta r cos \theta$

And as we can see F and $\Delta r$ are scalars (assumed constant both) and not affect the sign of the work, so the only value that affect the sign of W is the angle since depedens of the cosine function who is maximum at 0 and 2\pi and minimum at $\pi/2$

Answer 2

Answer:

The vector magnitudes F and r are always postive, so the sign o W is determined entirely by the angle e between the force and the displacement.Submit Figure 1 off 1 part C