How much work is done by the force F⃗ =(− 3.0 i^− 7.6 j^)N on a particle that moves through displacement Δr⃗ = 3.6 i^−2.0j^m?
Fx dx + Fy dy
(-3 * 3.6) - (7.6 * -2)
4.4
F=<-3,-7.6>
Δr = <3.6,-2>
Work done
F.Δr
= <-3,-7.6><3.6,-2>
= 4.4 J
Well, since work is calculated using the dot product of the force and displacement vectors, we have:
Work = F⃗ ⋅ Δr⃗
So, let's calculate that:
F⃗ ⋅ Δr⃗ = (− 3.0 i^− 7.6 j^) ⋅ (3.6 i^−2.0j^)
Now, to find the dot product, we multiply the corresponding components and add them up:
Work = (− 3.0 × 3.6) + (− 7.6 × −2.0)
Work = (− 10.8) + (15.2)
Work = 4.4
Therefore, the work done by the force F⃗ on the particle is 4.4 Joules. And here's a little joke for you: Why did the particle bring a pencil to the physics exam? Because it wanted to sketch out its work!
To calculate the work done by a force on a particle, you can use the formula:
Work (W) = Force (F) * Displacement (Δr) * cos(θ)
where θ is the angle between the force vector and the displacement vector.
In this case, we are given the force vector F⃗ = (-3.0 i^ - 7.6 j^) N and the displacement vector Δr⃗ = (3.6 i^ - 2.0 j^) m.
First, we need to find the magnitude of the force vector F:
|F| = sqrt((-3.0)^2 + (-7.6)^2) = sqrt(9 + 57.76) = sqrt(66.76) ≈ 8.17 N
Next, we need to find the angle between the force vector and the displacement vector. We can use the dot product to find the cosine of the angle:
cos(θ) = (F⃗ · Δr⃗) / (|F| * |Δr|)
(F⃗ · Δr⃗) = (-3.0 * 3.6) + (-7.6 * -2.0) = 10.8 + 15.2 = 26
|F| = 8.17 N (as calculated earlier)
|Δr| = sqrt((3.6)^2 + (-2.0)^2) = sqrt(12.96 + 4) = sqrt(16.96) ≈ 4.12
cos(θ) = 26 / (8.17 * 4.12) ≈ 0.785
Finally, we can calculate the work done by the force:
W = F * Δr * cos(θ) = 8.17 N * 4.12 m * 0.785 ≈ 26.84 J
Therefore, the work done by the force F⃗ on the particle is approximately 26.84 Joules.