A ball and a cube of equal mass lie on the floor. Both bodies are lifted up to the ceiling. In which case more work is done?

Answer 1

As we know, if the work is done by the force in the direction of displacement, it could be found using the formula

W = F·d.

Here F = mg - force against weight (we assume that in both cases we move the bodies with constant speed),

d - displacement (coincides with distance in case of motion along a straight line) of the center of mass of the body.

When both bodies are on the level of a floor, center of mass of cube is at lower level than the center of mass of the sphere.

Let's prove it. We assume that a ball and a cube are made from the same material (densities the same). Because they have the same mass, they have equal volumes V. Center of mass of a cube is at the level of h_c = a/2, where a - is the length of its edge. Center of mass of a sphere is at its geometric center - it is at height h_s = R above the floor level (R - radius of the sphere). For simplicity we will compare cubes of both heights above floor level.

For cube (2h_c ) ³ = a³ = V. Or (h_c ) ³ = V / 8 .

For sphere h_s³ = R ³ = 3V / 4π = V / 4.2.

It means that h_c < h_s. If the ceiling is at level H above the floor, then the center of mass of a cube we move through distance H - 2h_c. The center of mass of the sphere we move through distance H - 2h_s . Hence, we move cube's center of mass at greater distance. In both cases we apply the same force (masses of bodies the same). As a result, the work required to move the cube from the floor to the ceiling is greater than the work to move the sphere.

Answer: More work is done for cube