Question

A 2.20kg pendulum starts from a height of 5.00m . It swings back and forth through one whole oscillation but only returns to a maximum height of 4.75m . How much negative work was done on the pendulum during the first entire oscillation?

Answer 1

-5.396Joules

Step-by-step explanation:

Work is said to be done when a force causes a body to move through a distance and acts in the direction of the force.

Work done = Force × Distance

Given force = mg

Force = 2.20×9.81

Force applied on the string = 21.58N

Distance will be the difference in height = 4.75-5.00

= 0.25m

Negative work done on the pendulum during the first entirely oscillation = 21.58×0.25

= -5.396Joules

Answer 2

Negative workdone = 5.39J

Step-by-step explanation:

Mass (m) = 2.20kg

h₁ = 5.0m

h₂ = 4.75m

g = 9.8m/s²

Potential energy (PE) = mgh

M = mass

g = acceleration due to gravity

h = height

Work done = force * distance

Force = mass * acceleration (a or g)

Work done = mg * s

P.E = mgs = mgh

Negative potential energy of the pendulum = Mgh₂ - Mgh₁

P.E = mg(h₂-h₁)

P.E = 2.20 * 9.80 * (5.0 - 4.75)

P.E = 21.56 * 0.75

P.E = 5.39J

The negative potential work done = 5.39J