A mass m = 17.0 kg is pulled along a horizontal floor with NO friction for a distance d =5.1 m. Then the mass is pulled up an incline that makes an angle è = 30.0° with the horizontal and has a coefficient of kinetic friction ìk = 0.34. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of è = 30.0° (thus on the incline it is parallel to the surface) and has a tension T =66.0 N.

Incomplete.

To determine the work done on the mass, you need to calculate the work done on it in both the horizontal and inclined sections.

First, let's calculate the work done in the horizontal section:

Work done in the horizontal section (W1) is equal to the force applied (F1) multiplied by the distance covered (d1).

F1 = T (tension)
d1 = distance

Given:
T = 66.0 N
d1 = 5.1 m

W1 = F1 * d1
W1 = 66.0 N * 5.1 m
W1 = 336.6 Joules

Now, let's calculate the work done in the inclined section:

In the inclined section, we have two forces acting on the mass: the force of gravity (mg) acting vertically downward and the force of friction (fk) acting parallel to the incline in the opposite direction of the applied force.

The force of gravity can be broken down into two components: parallel to the incline (mg sin(theta)) and perpendicular to the incline (mg cos(theta)), where theta is the angle of the incline (30 degrees).

The net force acting on the mass in the direction of motion is:
F_net = T - fk

The work done in the inclined section (W2) is equal to the net force multiplied by the distance covered (d2).

fk = coefficient of kinetic friction * normal force
fk = μk * (mg cos(theta))

Given:
μk = 0.34 (coefficient of kinetic friction)
m = 17.0 kg (mass)
g = 9.8 m/s^2 (acceleration due to gravity)
theta = 30.0 degrees (angle of incline)
T = 66.0 N (tension)

Now, let's calculate fk first:

fk = μk * (mg cos(theta))
fk = 0.34 * (17.0 kg * 9.8 m/s^2 * cos(30.0 degrees))
fk = 0.34 * (166.6 N * 0.866)
fk = 0.34 * 144.1164 N
fk = 49.0202 N

Next, calculate the net force:

F_net = T - fk
F_net = 66.0 N - 49.0202 N
F_net = 16.9798 N

Finally, calculate the work done in the inclined section:

W2 = F_net * d2
d2 = distance

Given:
d2 = distance

To calculate d2, we use the geometry of the inclined plane. The distance along the incline is equal to d1 multiplied by the cosine of the angle of the incline.

d2 = d1 * cos(theta)
d2 = 5.1 m * cos(30.0 degrees)
d2 = 5.1 m * 0.866
d2 = 4.4116 m

W2 = F_net * d2
W2 = 16.9798 N * 4.4116 m
W2 = 74.9615 Joules

Therefore, the total work done on the mass is the sum of W1 and W2:

Total Work = W1 + W2
Total Work = 336.6 Joules + 74.9615 Joules
Total Work = 411.5615 Joules

So, the total work done on the mass is approximately 411.56 Joules.