Question

An object with mass M attached to the end of a string is raised vertically at a constant acceleration of g/8. If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?

a) (−9M g ℓ)/8
b) (−M g ℓ)/8
c) (M g ℓ)/4
d) (M g ℓ)
e) (9M g ℓ)/8
f) (M g ℓ)/8
g) (−M g ℓ)
h) (−M g ℓ)/4

Answer 1

The work done by the tension in the string when an object of mass M is raised vertically at a constant acceleration of g/8 for a distance l is (9Mgl)/8.

Step-by-step explanation:

To calculate the work done by the tension in the string when lifting an object with mass M at a constant acceleration of g/8, we need to consider two forces acting on the object: the gravitational force (Mg) and the tension in the string (T). The net force acting on the object is the sum of these two forces because the object is accelerating upwards. Since the acceleration is g/8, using Newton's second law, we get T - Mg = Ma where a = g/8.

To find T, we rearrange the equation to T = Mg + Ma = Mg + M(g/8) = Mg(1 + 1/8) = (9/8)Mg.

The work done by the tension (W) is the product of the tension force (T) and the distance (l) the object is raised: W = Tl = (9/8)Mgl. Thus, the work done by tension in the string is (9/8)Mgl.