Starting from rest, a body slides down a 45° incline. To solve this problem, we can break down the force acting on the body into its components.

First, let's label the forces acting on the body:

- Fg: Force due to gravity (weight of the body)
- Fn: Normal force (perpendicular to the incline)
- Ff: Force of friction (parallel to the incline)
- Fg//: Component of Fg parallel to the incline
- Fg⊥: Component of Fg perpendicular to the incline

Since the body is sliding down the incline, the force of friction is acting in the opposite direction to the motion. Thus, Ff = -μFn, where μ is the coefficient of friction.

To find the component of Fg parallel to the incline (Fg//), we can use trigonometry. The angle of the incline is given as 45°, so:

Fg// = Fg * sin(45°)

To find the component of Fg perpendicular to the incline (Fg⊥), we use trigonometry again:

Fg⊥ = Fg * cos(45°)

Since the body is starting from rest, we can assume that the force of friction is the only force acting to oppose the motion. Therefore, the net force acting on the body is:

Fnet = Ff = -μFn = -μ(Fg⊥)

Using Newton's second law (F = ma), we can relate the net force to the acceleration of the body:

Fnet = m * a

Since the body is sliding down the incline, the acceleration is in the direction of the incline. Therefore, the net force is equal to the component of Fg parallel to the incline:

Fnet = Fg// = Fg * sin(45°)

Setting these two equations equal to each other:

Fg * sin(45°) = -μ(Fg⊥)

Now, we can substitute the expressions for Fg// and Fg⊥:

Fg * sin(45°) = -μ(Fg * cos(45°))

Simplifying:

sin(45°) = -μ * cos(45°)

Since sin(45°) = cos(45°) = 1/√2, we have:

1/√2 = -μ * 1/√2

Simplifying further:

1 = -μ

Therefore, the coefficient of friction (μ) is equal to -1.

Note: This result is not physically meaningful since the coefficient of friction cannot be negative. It indicates that our assumption that the force of friction is the only force opposing the motion is incorrect, and there must be another force acting on the body.