A force ~F = Fx ˆı+Fy ˆ| acts on a particle that
undergoes a displacement of ~s = sx ˆı + sy ˆ|
where Fx = 10 N, Fy = −1 N, sx = 5 m, and
sy = 3 m.
Find the work done by the force on the
particle.
Answer in units of J
b)
Find the angle between ~F and ~s.
Answer in units of �
To find the work done by the force on the particle, we can use the formula:
Work (W) = Force (F) * Displacement (d) * cos(theta)
In this case, the force is given by:
~F = Fx ˆı + Fy ˆ|
where Fx = 10 N and Fy = -1 N.
The displacement ~s is given by:
~s = sx ˆı + sy ˆ|
where sx = 5 m and sy = 3 m.
Step 1: Calculate the dot product of the force vector ~F and the displacement vector ~s.
Dot Product = Fx * sx + Fy * sy
Dot Product = (10 N * 5 m) + (-1 N * 3 m)
Dot Product = (50 Nm) + (-3 Nm)
Dot Product = 47 Nm
Step 2: Calculate the magnitude of the force F.
Magnitude of F = sqrt(Fx^2 + Fy^2)
Magnitude of F = sqrt((10 N)^2 + (-1 N)^2)
Magnitude of F = sqrt(100 N^2 + 1 N^2)
Magnitude of F = sqrt(101 N^2)
Magnitude of F ≈ 10.05 N
Step 3: Calculate the angle between ~F and ~s.
cos(theta) = Dot Product / (Magnitude of F * Magnitude of s)
cos(theta) = 47 Nm / (10.05 N * sqrt((5 m)^2 + (3 m)^2))
cos(theta) = 47 Nm / (10.05 N * sqrt(34 m^2))
cos(theta) ≈ 0.469
Step 4: Calculate the work done by the force on the particle.
Work (W) = Force * Displacement * cos(theta)
Work (W) = (10.05 N) * (sqrt((5 m)^2 + (3 m)^2)) * 0.469
Work (W) = 10.05 N * (sqrt(34 m^2)) * 0.469
Work (W) ≈ 60.21 J (Answer to part a)
The work done by the force on the particle is approximately 60.21 J.
To find the angle between ~F and ~s, we can use the formula:
cos(theta) = Dot Product / (Magnitude of F * Magnitude of s)
Rearranging the formula, we have:
theta = acos(Dot Product / (Magnitude of F * Magnitude of s))
Substituting the values we found earlier:
theta = acos(47 Nm / (10.05 N * sqrt((5 m)^2 + (3 m)^2)))
theta = acos(47 Nm / (10.05 N * sqrt(34 m^2)))
theta ≈ 62.45 degrees (Answer to part b)
The angle between ~F and ~s is approximately 62.45 degrees.