A 0.75-kg block slides on a frictionless surface with a speed of 2 m/s. It then slides over a rough area 1.0 m in length and onto another frictionless surface. The coefficient of kinetic friction between the block and the rough surface is 0.17. What is the speed of the block after it passes across the rough surface

the speed is 0.88m/s

Explanation:

The mass is 0.75kg, and the velocity is 2m/s.

Now, for a distance of 1 meter, the block experiences a force of friction, that can be described by:

F = m*g*k, that is in the opposite direction to the movement of the block, and where k is the coefficient of kinetic friction, so we have that the force is:

F = -0.75*9.8*0.17N = -1.2N

Then the acceleration of the block is:

a = F/m = -1.2/0.75 m/s^2 = -1.6m/s^2

Now, the velocity equation will be:

v(t) = (-1.6m/s^2)*t + 2m/s

the position equation will be:

p(t) = (1/2)*(-1.6 m/s^2)*t^2 + 2m/s*t

when p(t) = 1m, the force of friction stops, so we must find this time and then put it in the equation of the velocity:

1 = (-0.8)*t^2 + 2*t

-0.8t^2 + 2t - 1 = 0

the roots can be obtained by:

t = (-2 +- √(2*2 - 4*-1*-0.8))/2*(-0.8) = (-2 +- 0.89)/(-1.6)

then the roots are:

t = (-2 + 0.89)/(-1.6) = 0.7s

t =  (-2 - 0.89)/(-1.6) = 1.8s

Here we must choose the smallest positive number, because once the block gets out of the rough area, the acceleration stops.

now, we put that time in the velocity equation:

v(0.7s) = (-1.6m/s^2)*0.7 + 2m/s = 0.88m/s

where are the vectors ?

there are no contact forces mentioned on the list of choices that you've provided.