A force F⃗ = 12 N i^ − 10 N j^ acts on an object. How much work does this force do as the object moves from the origin to the point r⃗ = 13 m i^ + 11 m j^?
W = F • d = Fx•rx + Fy•ry = (12 N) (13 m) + (−1 0 N)(+11 m) = 156 -110=46 J
Well, it seems like this force is on a magical adventure from the origin to a new point. Let's calculate the work done by this force!
The work done (W) by a force is given by the dot product of the force (F) and the displacement (Δr) of the object. So, let's find the displacement first.
The displacement (Δr) is the difference between the final position (r_final) and the initial position (r_initial). In this case, r_final = 13 m i^ + 11 m j^ and r_initial = 0 m i^ + 0 m j^.
Δr = r_final - r_initial
= (13 m i^ + 11 m j^) - (0 m i^ + 0 m j^)
= 13 m i^ + 11 m j^
Now, we can calculate the work done by taking the dot product of the force and the displacement.
W = F⃗ · Δr
= (12 N i^ - 10 N j^) · (13 m i^ + 11 m j^)
= (12 N * 13 m) + (-10 N * 11 m)
= 156 Nm - 110 Nm
= 46 Nm
So, the work done by this force as the object moves from the origin to the given point is 46 Nm.
To find the work done by the force as the object moves from the origin to point r⃗, we can use the equation:
Work = Force * Displacement * cos(θ)
Where:
- Force (F⃗) is the given force vector,
- Displacement is the displacement vector from the origin to the point r⃗, and
- θ is the angle between the force vector and the displacement vector.
First, we need to calculate the displacement vector:
Displacement (r⃗) = Final position vector (r⃗) - Initial position vector (r⃗₀)
Given:
r⃗ = 13 m i^ + 11 m j^ (Final position)
r⃗₀ = 0 m i^ + 0 m j^ (Initial position)
Therefore,
Displacement (r⃗) = (13 m i^ + 11 m j^) - (0 m i^ + 0 m j^)
= 13 m i^ + 11 m j^
Next, we need to calculate the angle (θ) between the force vector (F⃗) and the displacement vector (r⃗).
The dot product of two vectors is given by:
A · B = |A| |B| cos(θ)
We can rearrange the equation to solve for cos(θ):
cos(θ) = (A · B) / (|A| |B|)
In our case, A = F⃗ = 12 N i^ − 10 N j^ and B = r⃗ = 13 m i^ + 11 m j^.
Calculating the dot product:
A · B = (12 N i^ − 10 N j^) · (13 m i^ + 11 m j^)
= (12 N)(13 m) + (-10 N)(11 m)
= 156 N·m - 110 N·m
= 46 N·m
Calculating the magnitudes of A and B:
|A| = √((12 N)^2 + (-10 N)^2) = √(144 N^2 + 100 N^2) = √(244 N^2) = √(244) N ≈ 15.62 N
|B| = √((13 m)^2 + (11 m)^2) = √(169 m^2 + 121 m^2) = √(290 m^2) = √(290) m ≈ 17.03 m
Now, we can calculate cos(θ):
cos(θ) = (A · B) / (|A| |B|)
= (46 N·m) / (15.62 N * 17.03 m)
≈ 0.1602
Finally, we can calculate the work done:
Work = Force * Displacement * cos(θ)
= (12 N i^ − 10 N j^) * (13 m i^ + 11 m j^) * 0.1602
Multiplying the vectors component-wise:
Work = (12 N * 13 m) + (-10 N * 11 m) * 0.1602
= 156 N·m - 110 N·m * 0.1602
= 156 N·m - 17.6235 N·m
≈ 138.38 N·m
Therefore, the work done by the force as the object moves from the origin to the point r⃗ = 13 m i^ + 11 m j^ is approximately 138.38 N·m.
To find the work done by a force, we can use the formula:
Work = Force * Displacement * Cos(θ)
where:
- Force (F⃗) is the magnitude and direction of the force acting on the object,
- Displacement is the distance the object moves in the direction of the force, and
- θ is the angle between the force vector and the displacement vector.
In this case, we are given the force vector F⃗ = 12 N i^ - 10 N j^ and the displacement vector r⃗ = 13 m i^ + 11 m j^.
First, find the magnitude of the force using the Pythagorean theorem:
|F⃗ | = sqrt((12 N)^2 + (-10 N)^2) = sqrt(144 N^2 + 100 N^2) = sqrt(244 N^2) = 2 sqrt(61) N
Next, find the displacement vector's magnitude:
|r⃗ | = sqrt((13 m)^2 + (11 m)^2) = sqrt(169 m^2 + 121 m^2) = sqrt(290 m^2)
Now, calculate the dot product of the force and displacement vectors:
F⃗ · r⃗ = (12 N)(13 m) + (-10 N)(11 m) = 156 Nm - 110 Nm = 46 Nm
Then, calculate the cosine of the angle between the force and displacement vectors:
Cos(θ) = (F⃗ · r⃗) / (|F⃗ | * |r⃗ |) = (46 Nm) / (2 sqrt(61) N * sqrt(290) m) ≈ 0.374
Finally, substitute the values into the formula for work:
Work = (2 sqrt(61) N) * (sqrt(290) m) * 0.374
≈ 37.613 Nm
Therefore, the work done by the force is approximately 37.613 Nm.