Find the total work done by a 15 N force in the direction of the vector [1,2,2], when it moves a particle from O[0,0,0] to P[1,-3,4] and then from P to A[7,2,5]. Distance is measured in metres.
To find the total work done by a force in moving a particle along a particular path, we can use the dot product of the force and the displacement vector along that path.
First, let's calculate the work done in moving the particle from O to P.
1. Calculate the displacement vector, d1, from O to P:
d1 = P - O = [1, -3, 4] - [0, 0, 0] = [1, -3, 4]
2. Normalize the force vector, F, so that its magnitude is 15 N:
F_normalized = (15 / ||[1, 2, 2]||) * [1, 2, 2]
= (15 / √(1^2 + 2^2 + 2^2)) * [1, 2, 2]
= (15 / √9) * [1, 2, 2]
= 5 * [1, 2, 2]
= [5, 10, 10]
3. Calculate the dot product between the normalized force vector and displacement vector:
work1 = F_normalized · d1 = [5, 10, 10] · [1, -3, 4]
= (5 * 1) + (10 * -3) + (10 * 4)
= 5 - 30 + 40
= 15
Therefore, the work done in moving the particle from O to P is 15 Joules.
Now, let's calculate the work done in moving the particle from P to A.
1. Calculate the displacement vector, d2, from P to A:
d2 = A - P = [7, 2, 5] - [1, -3, 4] = [6, 5, 1]
2. Normalize the force vector, F, so that its magnitude is 15 N (Note: the force vector remains the same throughout):
F_normalized = [5, 10, 10]
3. Calculate the dot product between the normalized force vector and displacement vector:
work2 = F_normalized · d2 = [5, 10, 10] · [6, 5, 1]
= (5 * 6) + (10 * 5) + (10 * 1)
= 30 + 50 + 10
= 90
Therefore, the work done in moving the particle from P to A is 90 Joules.
To find the total work done by the force, we can simply sum up the individual works:
total work = work1 + work2 = 15 + 90 = 105 Joules
Therefore, the total work done by the 15 N force in the direction of the vector [1, 2, 2] in moving the particle from O to P and then from P to A is 105 Joules.