An object of mass 2 kg moving with a constant velocity of 10 m s⁻¹ encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?

Given, $m = 2\text{ kg}, u = 10\text{ ms}^{-1}$

Frictional force = 7 N

Additional opposing force = 3 N

Total opposing force:

$$F = 7 + 3 = 10\text{ N}$$

Using Newton's second law

$$F = ma$$

$$10 = 2a$$

$$a = 5\text{ ms}^{-2}$$

Since the force opposes the motion

$$a = -5\text{ ms}^{-2}$$

Now, using the equation,

$$v^2 = u^2 + 2as$$

Since the object comes to rest

$$v = 0$$

$$0 = (10)^2 + 2 \times (-5) \times s$$

$$0 = 100 - 10s$$

$$10s = 100$$

$$s = 10\text{ m}$$

Therefore, the object travels 10 m before coming to rest.