A force ~F = Fx ˆý+Fy ˆ| acts on a particle that undergoes a displacement of ~s = sx ˆý + sy ˆ| where Fx = 10 N, Fy = −4 N, sx = 6 m, and sy = 1 m.
1. Find the work done by the force on the particle.
Answer in units of J
2. Find the angle between ~F and ~s.
Answer in units of �degrees
(Fx*sx)+(Fy*sy)= 60-4= 56 <------
To find the work done by the force on the particle, we can use the formula:
Work = Force * Displacement * cos(theta)
Where:
- Force (F) is the magnitude of the force vector
- Displacement (s) is the magnitude of the displacement vector
- theta is the angle between the force and displacement vectors
1. Finding the work done by the force on the particle:
Given: Fx = 10 N, Fy = -4 N, sx = 6 m, sy = 1 m.
First, let's find the magnitude of the force vector:
|F| = sqrt(Fx^2 + Fy^2)
= sqrt((10 N)^2 + (-4 N)^2)
= sqrt(100 N^2 + 16 N^2)
= sqrt(116 N^2)
= 10.77 N (approx)
Next, let's find the magnitude of the displacement vector:
|s| = sqrt(sx^2 + sy^2)
= sqrt((6 m)^2 + (1 m)^2)
= sqrt(36 m^2 + 1 m^2)
= sqrt(37 m^2)
= 6.08 m (approx)
Now, let's find the angle between the force and displacement vectors:
tan(theta) = Fy / Fx
theta = atan(Fy / Fx)
= atan((-4 N) / (10 N))
= atan(-0.4)
= -21.8 degrees (approx)
Note: We use atan function to find the angle in radians. To convert it into degrees, multiply the result by (180 / pi).
Finally, let's calculate the work done by the force:
Work = |F| * |s| * cos(theta)
= (10.77 N) * (6.08 m) * cos(-21.8 degrees)
= 64.85 J (approx)
Therefore, the work done by the force on the particle is 64.85 J.
2. Finding the angle between the force and displacement vectors:
We have already calculated the angle between the force and displacement vectors in the previous step. The angle is approximately -21.8 degrees.
To find the work done by a force on a particle, we can use the formula:
Work = Force * Displacement * cos(theta)
where Force and Displacement are both vectors, and theta is the angle between them.
1. Let's first calculate the work done by the force on the particle.
Given:
Force, ~F = Fx ˆý + Fy ˆ|
Displacement, ~s = sx ˆý + sy ˆ|
We can find the dot product of the force and displacement vectors:
~F · ~s = (Fx * sx) + (Fy * sy)
Substituting the given values:
~F · ~s = (10 N * 6 m) + (-4 N * 1 m)
= 60 N*m + (-4 N*m)
= 56 N*m
Therefore, the work done by the force on the particle is 56 N*m. Since work is measured in joules (J), the answer is 56 J.
2. Now let's find the angle between the force vector, ~F, and the displacement vector, ~s.
Using the dot product formula, we have:
~F · ~s = |~F| |~s| cos(theta)
Given:
~F = Fx ˆý + Fy ˆ|
~s = sx ˆý + sy ˆ|
The magnitudes of the vectors are:
|~F| = sqrt(Fx^2 + Fy^2)
= sqrt((10 N)^2 + (-4 N)^2)
= sqrt(100 N^2 + 16 N^2)
= sqrt(116 N^2)
= 10.77 N
|~s| = sqrt(sx^2 + sy^2)
= sqrt((6 m)^2 + (1 m)^2)
= sqrt(36 m^2 + 1 m^2)
= sqrt(37 m^2)
= 6.08 m
Plugging the values into the dot product formula:
(10.77 N * 6.08 m) cos(theta) = 56 N*m
Solving for cos(theta):
cos(theta) = (56 N*m) / (10.77 N * 6.08 m)
= 0.867
To find the angle, theta, we can take the inverse cosine (cos^(-1)) of 0.867 using a calculator.
theta ≈ 30.98 degrees
Therefore, the angle between ~F and ~s is approximately 30.98 degrees.