Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 55.3 kg, down a theta= 79.6º slope at constant acceleration a=-4.3 m/s2, as shown in Figure (here we assume the positive direction is going down the slope. So the given acceleration is a negative value, it means its direction is going up the slope, slowing down as it moving downward). So, the coefficient of friction between the sled and the snow is 0.100. How many Joules of work is done by the tension in the rope as the sled moves 2.1 m along the hill? Use g= 10 m/s2.

The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.

In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.

Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.

From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:

(1)

Where:

- Work done by tension, in joules.

- Mass of the sled-victim system, in kilograms.

- Coefficient of kinetic friction, no unit.

- Gravitational acceleration, in meters per square second.

- Travelled distance, in meters.

- Slope angle, in sexagesimal degrees.

- Net acceleration of the sled-victim system, in meters per square second.

If we know that , , , , and , then the work done by the tension in the rope is:

The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.

Related questions:

it is the second option.

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.