A 5 N force which acts along the direction vector (2, 3), moves an object from A(1, 3) to B(3, 7). Find the work done if the given units are metres.
Well, it seems like our object took a nice little stroll from point A to point B. Let's calculate the work done using the equation W = F * d * cos(theta).
First, we need to find the displacement vector. The displacement vector is the difference between the final position (B) and the initial position (A). So, our displacement vector is (3 - 1, 7 - 3), which simplifies to (2, 4).
Now, let's find the magnitude of the displacement vector, which is the distance between A and B. We can use the distance formula: sqrt((x2 - x1)^2 + (y2 - y1)^2). In this case, it's sqrt((3 - 1)^2 + (7 - 3)^2), which simplifies to sqrt(2^2 + 4^2), which further simplifies to sqrt(4 + 16), which finally gives us sqrt(20) = 2sqrt(5).
Okay, now we know our force is 5 N and the displacement is 2sqrt(5) meters. The angle between the force vector and the displacement vector is the angle between (2, 3) and (2, 4). To find this angle, we can use the dot product formula: dot_product_v1_v2 = |v1| * |v2| * cos(theta).
In this case, the dot product is (2 * 2) + (3 * 4) = 4 + 12 = 16. The magnitude of vector v1 is sqrt(2^2 + 3^2) = sqrt(4 + 9) = sqrt(13). The magnitude of vector v2 is 2sqrt(5). So, we have 16 = sqrt(13) * 2sqrt(5) * cos(theta).
Now, solve for cos(theta) by dividing both sides by sqrt(13) * 2sqrt(5). We get cos(theta) = 16 / (sqrt(13) * 2sqrt(5)), which simplifies to cos(theta) = 8 / (sqrt(13) * sqrt(5)), or cos(theta) = 8sqrt(5) / (sqrt(13) * sqrt(5)).
Finally, plug everything into the work formula: W = 5 * 2sqrt(5) * (8sqrt(5) / (sqrt(13) * sqrt(5))). Sqrt(5) cancels out, and we're left with W = 5 * 2 * 8 / sqrt(13), which simplifies to W = 80 / sqrt(13) meters. And there you have it, the work done is 80 / sqrt(13) meters.
To find the work done by a force, we need to calculate the dot product of the force vector and the displacement vector.
First, let's find the displacement vector from A to B:
Displacement vector = B - A = (3, 7) - (1, 3) = (2, 4)
Next, let's normalize the direction vector:
Direction vector = (2, 3) / ||(2, 3)||, where ||(2, 3)|| is the magnitude of (2, 3).
Magnitude of (2, 3) = sqrt(2^2 + 3^2) = sqrt(4 + 9) = sqrt(13)
Direction vector = (2, 3) / sqrt(13)
Now, let's calculate the dot product between the force and displacement vectors:
Work done = Force * displacement * cos(theta), where theta is the angle between the force and displacement vectors.
Force = 5 N
Displacement = ||(2, 4)|| = sqrt(2^2 + 4^2) = sqrt(4 + 16) = sqrt(20) meters (since the given units are meters)
Theta = angle between the force and displacement vectors
We can find cos(theta) using the dot product:
cos(theta) = (Force vector dot Product vector) / (||Force vector|| * ||Product vector||)
Force vector = (2, 3)
Product vector = (2, 4)
cos(theta) = (2 * 2 + 3 * 4) / (sqrt(2^2 + 3^2) * sqrt(2^2 + 4^2))
= (4 + 12) / (sqrt(4 + 9) * sqrt(4 + 16))
= 16 / (sqrt(13) * sqrt(20))
= 16 / (sqrt(13) * 2 * sqrt(5))
= 8 / (sqrt(13) * sqrt(5))
= 8 / (sqrt(65))
Now we can calculate the work done:
Work done = 5 * sqrt(20) * (8 / (sqrt(65)))
= 5 * 4 * sqrt(20) / (sqrt(65))
= 20 * (2 * sqrt(5)) / (sqrt(65))
= 40 * sqrt(5) / (sqrt(65))
= (40 * sqrt(5) * sqrt(65)) / (sqrt(65) * sqrt(65))
= (40 * sqrt(325)) / 65
= (8 * sqrt(325)) / 13
Therefore, the work done by the 5 N force is (8 * sqrt(325)) / 13 Joules.
To find the work done, we can use the formula:
Work = Force * Displacement * cos(theta)
Where:
- Force is the magnitude of the force vector
- Displacement is the magnitude of the displacement vector
- theta is the angle between the force and displacement vectors
First, let's find the magnitude of the force vector:
Magnitude of force = sqrt(2^2 + 3^2) = sqrt(4 + 9) = sqrt(13) N
Next, let's find the displacement vector:
Displacement vector = B - A = (3, 7) - (1, 3) = (2, 4)
Magnitude of displacement = sqrt(2^2 + 4^2) = sqrt(4 + 16) = sqrt(20) m
Next, let's calculate the angle between the force and displacement vectors:
cos(theta) = (force dot displacement) / (|force| * |displacement|)
force dot displacement = (2 * 2) + (3 * 4) = 4 + 12 = 16
cos(theta) = 16 / (sqrt(13) * sqrt(20))
Finally, let's calculate the work done:
Work = sqrt(13) * sqrt(20) * (16 / (sqrt(13) * sqrt(20)))
Simplifying the expression:
Work = 16
Therefore, the work done by the 5 N force in moving the object from A to B is 16 Joules.