We can use the equation of motion, `v^2 = u^2 + 2as`

where u is the initial velocity of the object, v is the final velocity of the object, s is the distance traveled and a is the acceleration.

In this case, the initial velocity of the car is 10...

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We can use the equation of motion, `v^2 = u^2 + 2as`

where u is the initial velocity of the object, v is the final velocity of the object, s is the distance traveled and a is the acceleration.

In this case, the initial velocity of the car is 10 m/s and it travels for 8 m.

Thus, u = 10 m/s and s = 8 m

Since the car stops, v = 0

Substituting these values, we get

`0 = 10^2 + 2a(8)`

Solving this equation, we get, a = - 6.25 m/s^2 (negative sign indicates deceleration).

Using Newton's Second Law of Motion, force is related to mass and acceleration of an object as F = ma.

Here, the mass of the car m = 800 kg

Thus, the average force exerted by the brakes = ma = 800 x 6.25 = **5,000 N**.

Hope this helps.