A cat of mass m climbs over an inclined plane of height h and then climbs down along the vertical side of the same inclined plane until it reaches the base of the said inclined plane. Calculate the work made by the gravitational force applied onto the cat by the Earth a) while climbing; b) while climbing down; c) total work.

19.62mh or 2mgh

Explanation:

the mass of the cat = m

height of the inclined plane = h

NB: The work made by the gravitational force applied onto the cat by the Earth will be equal to the work done by the cat to overcome this force.

a) For climbing (directly) up the inclined plane, the work done by the cat against gravity will be

work = weight x distance

weight = mg = mass x acceleration due to gravity

where g acceleration due to gravity = 9.98 m/s^2

work = -mgh = -9.81mh

the negative sign indicates that the work is done against gravity

b) while climbing down the vertical side of the inclined plane, work done by the cat is the same amount of work the cat does in climbing (directly) up the inclined plane, if we disregard losses due to friction. The only advantage of the inclined plane is that the inclined plane allows the same work to be done with a smaller force exerted over a greater distance; by spreading the distance (now the length of the inclined plane), and reducing the force (weight of the cat) by resolving it into a smaller force that is normal to the inclined plane. All in all the work done is equal and is in accordance with the conservation of energy.

the work done by gravity on the cat on climbing down is

work = mgh = 9.81mh

c) total work done = final work - initial work = 9.81mh - (-9.81mh) = 19.62mh

the components of a vector are the parts of the vectors that are to each other.

b perpendicular