Question

A 14 N force with fixed orientation does work on particle as the particle moves through displacement d = 3i ₅j+5k the displacement if the change in the particle's kinetic energy is (a) +10.8 and (b) -10.8 J? What is the angle between the force and (a) Number Units (b) Number Units.

Answer 1

(a) The angle between the force and the displacement is θ = 60 degrees.

(b) The angle between the force and the displacement is θ = 120 degrees.

Step-by-step explanation:

In order to find the angle θ between the force and the displacement, we can use the formula for work done (W) by a constant force:

`\[ W = \vec{F} \cdot \vec{d} \cdot \cos(\theta) \]`

where
`\(\vec{F}\)` is the force,
`\(\vec{d}\)` is the displacement vector, and
`\(\theta\)` is the angle between the force and the displacement. The work done is also equal to the change in kinetic energy (ΔKE).

Given that the work done (ΔKE) is +10.8 J in case (a) and -10.8 J in case (b), we substitute these values into the equation:

(a)
`\[ +10.8 = 14 \cdot 3 \cdot \cos(\theta) \]`

(b)
`\[ -10.8 = 14 \cdot 3 \cdot \cos(\theta) \]`

Solving for
`\(\theta\)` in each case, we find:

(a)
`\[ \cos(\theta) = (10.8)/(42) \]`

(b)
`\[ \cos(\theta) = (-10.8)/(42) \]`

Taking the inverse cosine to find the angle, we get:

(a)
`\[ \theta = \cos^(-1)\left((10.8)/(42)\right) \approx 60^\circ \]`

(b)
`\[ \theta = \cos^(-1)\left((-10.8)/(42)\right) \approx 120^\circ \]`

Therefore, the angles between the force and displacement for (a) and (b) are 60 degrees and 120 degrees, respectively.