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.

Work done by a force F along a vector P is F.P (dot product). So total work done is

F.P + F.A
where A=<1,-3,4>, P=<7-1, 2-(-3), 5-4>
and F=15<1,2,2>/sqrt(1^2+2^2+2^2)=<5,10,10>
The division by sqrt(...) is required to normalize the direction vector to a unit vector.

I don't understand it.. what is the final work done? Also, the arrows (< and >) are they actually supposed to be written as < and > or are they supposed to mean brackets ( ) or these brackets []

thank you

Most books write vectors as <x,y,z> to differentiate from a point (x,y,z).

You would separate the work done by P along the two lines (vectors) OP and PA.

The force is 15N along <1,2,2>. Since <1,2,2> is not a unit vector, we need to normalize it as
<1,2,2>/sqrt(1^2+2^2+2^2)=<1,2,2>/3
So the force is
F=15<1,2,2>/3 = 5<1,2,2>=<5,10,10>

Work done is F.D, where F and D are vectors, and . represents the dot product.

The first part of the work is
F.OP=<5,10,10>.<1-0,-3-0,4-0>
=<5,10,10>.<1,-3,4>
=5*1-10*3+10*4
=15

I'll leave it to you to calculate the second part, F.PA, where
A=<7-1, 2-(-3), 5-4>=<6,5,1>

could you help me calculate the second part also? thank you

F.PA=<5,10,10><6,5,1>

=5*6+10*5+10*1
=30+50+1
=81

Isnt it 30 + 50 + 10 because 10* 1 is 10 so final is = 90

To find the total work done by a force, we use the formula:

Work = Force * Distance * cos(theta),

where
- Work is the total work done by the force,
- Force is the magnitude of the force,
- Distance is the distance through which the force is applied,
- theta is the angle between the force vector and the direction of displacement.

Let's break down the problem into two parts: O to P and P to A. Then we can calculate the work done for each part and add them together to get the total work done.

1. O to P:

- Force: Given as 15 N.
- Distance: To find the distance between O and P, we can use the formula for Euclidean distance in three dimensions:

Distance(OP) = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2),

where (x1, y1, z1) are the coordinates of point O and (x2, y2, z2) are the coordinates of point P.

Plugging in the coordinates, we get:

Distance(OP) = sqrt((1 - 0)^2 + (-3 - 0)^2 + (4 - 0)^2)
= sqrt(1 + 9 + 16)
= sqrt(26) meters.

- Theta: To find the angle theta between the force vector [1, 2, 2] and the displacement vector from O to P, we can use the dot product formula:

cos(theta) = (A · B) / (|A| |B|),

where A and B are the given vectors.

Plugging in the values, we get:

cos(theta) = ([1, 2, 2] · [1, -3, 4]) / (sqrt(1^2 + 2^2 + 2^2) * sqrt(1^2 + (-3)^2 + 4^2))
= (1 - 6 + 8) / (sqrt(1 + 4 + 4) * sqrt(1 + 9 + 16))
= 3 / (sqrt(9) * sqrt(26))
= 3 / (3 * sqrt(26))
= 1 / sqrt(26).

Now, we can calculate the work done for O to P:

Work(OP) = Force * Distance(OP) * cos(theta)
= 15 N * sqrt(26) meters * (1 / sqrt(26))
= 15 N * sqrt(26) meters
≈ 76.491 N·m (rounded to three decimal places).

2. P to A:

- Force: Again, given as 15 N.
- Distance: We need to find the distance between P and A using the same Euclidean distance formula:

Distance(PA) = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2),

where (x1, y1, z1) are the coordinates of point P and (x2, y2, z2) are the coordinates of point A.

Plugging in the coordinates, we get:

Distance(PA) = sqrt((7 - 1)^2 + (2 - (-3))^2 + (5 - 4)^2)
= sqrt(36 + 25 + 1)
= sqrt(62) meters.

- Theta: We need to find the angle theta between the force vector [1, 2, 2] and the displacement vector from P to A. Again, we use the dot product formula:

cos(theta) = (A · B) / (|A| |B|),

where A and B are the given vectors.

Plugging in the values, we get:

cos(theta) = ([1, 2, 2] · [7, 2, 5]) / (sqrt(1^2 + 2^2 + 2^2) * sqrt(7^2 + 2^2 + 5^2))
= (7 + 4 + 10) / (sqrt(9) * sqrt(74))
= 21 / (3 * sqrt(74))
= 7 / sqrt(74).

Now, we calculate the work done for P to A:

Work(PA) = Force * Distance(PA) * cos(theta)
= 15 N * sqrt(62) meters * (7 / sqrt(74))
= 15 N * 7 * sqrt(62 / 74) meters
≈ 74.409 N·m (rounded to three decimal places).

Finally, we can find the total work done by summing up the work done for O to P and P to A:

Total Work = Work(OP) + Work(PA)
≈ 76.491 N·m + 74.409 N·m
≈ 150.900 N·m (rounded to three decimal places).

Therefore, the total work done by the 15 N force in the direction of the vector [1, 2, 2], when moving the particle from O to P and then from P to A, is approximately 150.900 N·m.