A cord is used to lower vertically a block of mass \(\mathrm{M}\) by a distance \(\mathrm{d}\) with constant downward acceleration \((9 / 2)\). Work done by the cord on the block is (A) \(-\mathrm{Mgd} / 2\) (B) \(\mathrm{Mgd} / 4\) (C) \(-3 \mathrm{Mgd} / 4\) (D) \(\mathrm{Mgd}\)

Let's identify the forces acting on the block: 1. Gravitational force (downward) is equal to Mg. 2. Tension force in the cord acts upward. Since the block is having constant downward acceleration of (9/2), the net force (F_net) acting on the block is also downward.

Using Newton's second law for the motion of the block, we have \(F_{net} = Ma\), where M is the mass of the block, and a is the downward acceleration. So, \(F_{net} = M \times \frac{9}{2}\)

To find the tension force (T) in the cord, we'll use the following equation: \(F_{net} = T - F_g\) Where \(F_g\) is the gravitational force (Mg) acting on the block. Replacing the net force and the gravitational force in the equation, we get: \(M \times \frac{9}{2} = T - Mg\) Solving for T, we get: \(T = M \times (\frac{9}{2} + g)\)

By definition, work (W) is given by the equation \(W = Fd\cos\theta\), where F is the force applied, d is the displacement, and \(\theta\) is the angle between the force and displacement. Since the tension in the cord opposes the displacement of the block (acts upwards), the angle between the tension force and the displacement is 180 degrees. Therefore, the value of \(\cos\theta = -1\). To find the work done by the cord (W_cord), we'll use tension force (T) found in step 3: \(W_{cord} = T \times d \times (-1)\) \(W_{cord} = -M \times (\frac{9}{2} + g) \times d\) Substituting the value of g: \(W_{cord} = -M \times (\frac{9}{2} + 9) \times d\) Simplifying the equation, we get: \(W_{cord} = -\frac{Mgd}{2}\) The work done by the cord on the block is option (A) \(-\frac{Mgd}{2}\).