A force F=(5i+3j)N acts on a body and causes a displacement r=(7i-j)m. Determine the work done?

work is done by the x component of force moving in x direction and by the y component moving in y direction

(5*7) + (3*-1)
= 35 - 3
= 32 Joules

note the -3 means the force is opposite to the motion (braking, work out)

your text probably expressed the work as Fr

To determine the work done, we use the formula:

Work = Force dot Product Displacement

The dot product of two vectors is given by the formula:

A dot B = Ax * Bx + Ay * By + Az * Bz

In this case, we have:

Force F = 5i + 3j
Displacement r = 7i - j

Let's calculate the dot product of these two vectors:

F dot Product r = (5 * 7) + (3 * -1)
= 35 - 3
= 32

Therefore, the work done is 32 Joules.

To determine the work done, you need to use the formula:

Work = Force * Displacement * cos(theta)

where:
- Force is the magnitude and direction of the force vector
- Displacement is the magnitude and direction of the displacement vector
- theta is the angle between the force and displacement vectors

In this case, we have:
- Force F = 5i + 3j N
- Displacement r = 7i - j m

First, let's find the magnitudes of the force and displacement vectors:

Magnitude of Force F = sqrt((5)^2 + (3)^2) = sqrt(34) N

Magnitude of Displacement r = sqrt((7)^2 + (-1)^2) = sqrt(50) m

Next, we need to calculate the angle between the force and displacement vectors (theta). The angle between two vectors can be found using the dot product formula:

theta = arccos((F dot r) / (|F| * |r|))

The dot product of two vectors A and B is given by A dot B = (Ax * Bx) + (Ay * By)

In this case:

F dot r = (5 * 7) + (3 * -1) = 34

theta = arccos(34 / (sqrt(34) * sqrt(50)))

Using a calculator, we find that theta ≈ 41.4 degrees.

Finally, we can substitute the values into the work formula:

Work = (sqrt(34) N) * (sqrt(50) m) * cos(41.4 degrees)

Now, simply evaluate this expression to find the work done.