Question

An object with mass M is attached to the end of a string and is being lowered vertically at a constant acceleration of g 4 . If it has been lowered a distance ℓ from rest, how much work has been done by the tension in the string?

Answer 1

- 3 Mgl / 4

Step-by-step explanation:

Let T be the tension in the string.

mass of the object = M

Downward acceleration, a = g / 4

By Newton's second law

Mg - T = Ma

T = Mg - Ma

T = M ( g - g/4)

T = 3Mg/4

Distance = l

Work done by the tension

W = T x d x cos 180

Because the tension is upward and the distance if downward, so the angle between them is 180°.

W = T x d x (-1)

W = - 3Mgl/4

Thus, the work done is - 3 Mgl / 4.