A cord is used to vertically lower an initially stationary block of mass M at a constant downward acceleration of g/4. When the block has fallen a distance d, find (a) the work done by the cord's force on the block, (b) the work done by the graviational force on the block, (c) the kinitic energy of the block, and (d) the speed of the block.

Answer

To solve the problem step by step, we will analyze the forces acting on the block and calculate the required quantities.

### Given:
- Mass of the block = \( M \)
- Downward acceleration of the block = \( \frac{g}{4} \)
- Distance fallen = \( d \)

### Step 1: Find the tension in the cord (T)

Using Newton's second law, we can analyze the forces acting on the block:

1. The gravitational force acting downward is \( F_g = Mg \).
2. The tension in the cord acts upward, denoted as \( T \).

Since the block is accelerating downward, we can write the equation of motion as:

\[ F_{\text{net}} = M \cdot a \]

Where \( a = \frac{g}{4} \). Therefore, we have:

\[ Mg - T = M \cdot \frac{g}{4} \]

Rearranging gives:

\[ T = Mg - M \cdot \frac{g}{4} = Mg \left(1 - \frac{1}{4}\right) = Mg \cdot \frac{3}{4} \]

### Step 2: Calculate the work done by the cord's force on the block (W_T)

The work done by the tension force (cord's force) is given by:

\[ W_T = T \cdot d \cdot \cos(180^\circ) \]

Since the angle between the tension (upward) and the displacement (downward) is \( 180^\circ \):

\[ \cos(180^\circ) = -1 \]

Thus, the work done by the tension is:

\[ W_T = \left(\frac{3}{4}Mg\right) \cdot d \cdot (-1) = -\frac{3}{4}Mgd \]

### Step 3: Calculate the work done by the gravitational force on the block (W_g)

The work done by the gravitational force is given by:

\[ W_g = F_g \cdot d \cdot \cos(0^\circ) \]

Where \( F_g = Mg \) and the angle is \( 0^\circ \):

\[ W_g = Mg \cdot d \cdot 1 = Mgd \]

### Step 4: Calculate the kinetic energy of the block (K.E.)

Using the work-energy theorem, the total work done on the block is equal to the change in kinetic energy:

\[ W_{\text{total}} = W_T + W_g = K.E. - K.E._{\text{initial}} \]

Since the block starts from rest, \( K.E._{\text{initial}} = 0 \):

\[ W_{\text{total}} = -\frac{3}{4}Mgd + Mgd \]

Combining the terms:

\[ W_{\text{total}} = \left(1 - \frac{3}{4}\right)Mgd = \frac{1}{4}Mgd \]

Thus, the kinetic energy of the block when it has fallen a distance \( d \) is:

\[ K.E. = \frac{1}{4}Mgd \]

### Step 5: Calculate the speed of the block (v)

The kinetic energy is also given by the formula:

\[ K.E. = \frac{1}{2}Mv^2 \]

Setting the two expressions for kinetic energy equal gives:

\[ \frac{1}{4}Mgd = \frac{1}{2}Mv^2 \]

Dividing both sides by \( M \) (assuming \( M \neq 0 \)):

\[ \frac{1}{4}gd = \frac{1}{2}v^2 \]

Multiplying both sides by 2:

\[ \frac{1}{2}gd = v^2 \]

Taking the square root:

\[ v = \sqrt{\frac{gd}{2}} \]

### Summary of Results:
(a) Work done by the cord's force: \( W_T = -\frac{3}{4}Mgd \)
(b) Work done by the gravitational force: \( W_g = Mgd \)
(c) Kinetic energy of the block: \( K.E. = \frac{1}{4}Mgd \)
(d) Speed of the block: \( v = \sqrt{\frac{gd}{2}} \)