A force F~ = Fx ˆı + Fy ˆ acts on a particle that

undergoes a displacement of ~s = sx ˆı + sy ˆ
where Fx = 6 N, Fy = −1 N, sx = 6 m, and
sy = 3 m.
Find the work done by the force on the
particle.
Answer in units of J.
Find the angle between F~ and ~s.
Answer in units of ◦

Well, to find the work done by the force on the particle, we can use the formula: work = force * displacement * cos(theta). In this case, the force F~ = Fx ˆı + Fy ˆ, so we have F~ = (6 N)ˆı + (-1 N)ˆ. The displacement ~s = sx ˆı + sy ˆ, where sx = 6 m and sy = 3 m.

Now, let's calculate the work done by the force: work = (Fx*sx + Fy*sy) * cos(theta).

Substituting the given values, we have: work = (6 N * 6 m + (-1 N) * 3 m) * cos(theta).

Simplifying further, we get: work = (36 N * m - 3 N * m) * cos(theta).

That simplifies to: work = 33 N * m * cos(theta).

So, the work done by the force on the particle is 33 N * m * cos(theta).

To find the work done by the force on the particle, we can use the formula:

Work (W) = Force (F) * Displacement (s) * cos(theta)

Given:
Force Fx = 6 N
Force Fy = -1 N
Displacement sx = 6 m
Displacement sy = 3 m

To calculate the total force:
F = sqrt(Fx^2 + Fy^2)
= sqrt(6^2 + (-1)^2)
= sqrt(36 + 1)
= sqrt(37) N

To calculate the total displacement:
s = sqrt(sx^2 + sy^2)
= sqrt(6^2 + 3^2)
= sqrt(36 + 9)
= sqrt(45) m

To find the angle between F~ and ~s, we can use the formula:

cos(theta) = (Fx * sx + Fy * sy) / (F * s)

Substituting the given values:
cos(theta) = (6 * 6 + (-1) * 3) / (sqrt(37) * sqrt(45))
= (36 - 3) / (sqrt(37 * 45))
= 33 / sqrt(1665)

To find the value of theta, we need to find the inverse cosine (cos^-1) of cos(theta):

theta = cos^-1(33 / sqrt(1665))
≈ 83.7 degrees

Therefore, the work done by the force on the particle is approximately 6 * sqrt(37) * sqrt(45) * cos(theta) J and the angle between F~ and ~s is approximately 83.7 degrees.

To find the work done by the force on the particle, we can use the formula:

Work = Force dot Product Displacement

Where the dot product of two vectors is given by:

A • B = Ax * Bx + Ay * By

Given that F~ = Fx ˆı + Fy ˆ and ~s = sx ˆı + sy ˆ, we can substitute the given values:

F~ = 6 ˆı N - 1 ˆ N
~s = 6 ˆı m + 3 ˆ m

The dot product of F~ and ~s is:

F~ • ~s = (6 N)(6 m) + (-1 N)(3 m)

Now we can calculate the dot product:

F~ • ~s = 36 Nm - 3 Nm = 33 Nm

Therefore, the work done by the force on the particle is 33 Nm.

To find the angle between F~ and ~s, we can use the formula for the angle between two vectors:

cosθ = (A • B) / (|A| * |B|)

where cosθ is the cosine of the angle between the vectors A and B, and |A| and |B| are the magnitudes of vectors A and B.

In this case, A = F~ and B = ~s. The magnitudes of A and B are:

|A| = sqrt(Fx² + Fy²) = sqrt((6 N)² + (-1 N)²) = sqrt(36 N² + 1 N²) = sqrt(37) N
|B| = sqrt(sx² + sy²) = sqrt((6 m)² + (3 m)²) = sqrt(36 m² + 9 m²) = sqrt(45) m

Now we can substitute the given values:

cosθ = (33 Nm) / (sqrt(37) N * sqrt(45) m)

Simplifying this expression gives us:

cosθ ≈ 0.5941

To find the angle θ, we can take the inverse cosine (cos⁻¹) of 0.5941:

θ ≈ cos⁻¹(0.5941)

Evaluating this expression using a calculator gives:

θ ≈ 53.01°

Therefore, the angle between F~ and ~s is approximately 53.01°.

work done by Fx = 6 * 6 = 36 Joules

work done by Fy = -1 * 3 = -3 Joules
total work done = 33 Joules

tan F = -1/6 in quadrant 4 so -18.4 deg
tan s = 3/6 in quadrant 1 so 26.6
26.6 + 18.4 = 45 degrees