A sledge weighing 30 kg is pulled with a rope that makes 60° with the horizontal. The tension on the rope is constant and it is 37N. The sledge is moved horizontally for 13 m. There is no friction and cos(60°)=0.5

a) What is the work done by the tension on the sledge in Joules?
b) Define work.
c) The above was an example of a solid block doing work. Gases can also do work. Describe a situation where this is possible.

a) To find the work done by the tension on the sledge, we can use the formula:

Work = Force × Distance × cos(θ)

Here, the force is the tension on the rope, which is 37N, the distance is 13m, and θ is the angle between the force and displacement, which is 60°.

Work = 37N × 13m × cos(60°)
= 37N × 13m × 0.5
= 240.5 Joules

So, the work done by the tension on the sledge is 240.5 Joules.

b) Work is defined as the product of the force applied on an object and the displacement of the object in the direction of the force. It is a measure of how much energy is transferred to or from an object as a result of a force acting on it. Work is a scalar quantity and its units are Joules (J).

c) In the context of gases doing work, one example is the expansion or compression of a gas inside a cylinder with a movable piston. When the gas expands or gets compressed, it exerts a force on the piston. If the piston moves, the force applied by the gas does work on the surroundings. This work is done against external pressure and can be calculated using the formula:

Work = Pressure × Change in Volume

If the volume of the gas increases, the gas does work on the surroundings. Conversely, if the volume decreases, external forces do work on the gas. This concept is utilized in various applications such as engines, air compressors, and refrigeration systems.