A block of 200 g mass is dropped from a height of 2 m on to a spring and compress the spring to a distance of 50 cm. The force constant of the spring is

Options

A. `20 Nm^(-1)`
B. `40 Nm^(-1)`
C. `30 Nm^(-1)`
D. `60 Nm^(-1)`

Explanation

To find the force constant of the spring when a block of mass 200 g is dropped from a height of 2 m and compresses the spring by 50 cm, we can follow these steps:

### Step 1: Convert the mass to kilograms
The mass \( m \) is given as 200 g. To convert this to kilograms:
\[ m = \frac{200 \text{ g}}{1000} = 0.2 \text{ kg} \]

### Step 2: Convert the compression distance to meters
The compression of the spring \( x \) is given as 50 cm. To convert this to meters:
\[ x = \frac{50 \text{ cm}}{100} = 0.5 \text{ m} \]

### Step 3: Calculate the effective height
The effective height \( H \) from which the block falls is the sum of the height from which it is dropped and the compression of the spring:
\[ H = 2 \text{ m} + 0.5 \text{ m} = 2.5 \text{ m} \]

### Step 4: Use the conservation of energy principle
The potential energy (PE) of the block at height \( H \) is converted into the potential energy stored in the spring when it is compressed. The potential energy of the block is given by:
\[ PE = mgh \]
Substituting the values:
\[ PE = 0.2 \text{ kg} \times 10 \text{ m/s}^2 \times 2.5 \text{ m} = 5 \text{ J} \]

### Step 5: Set the potential energy equal to the spring potential energy
The potential energy stored in the spring when compressed is given by:
\[ PE_{spring} = \frac{1}{2} k x^2 \]
Setting the two potential energies equal:
\[ mgh = \frac{1}{2} k x^2 \]
Substituting the known values:
\[ 5 = \frac{1}{2} k (0.5)^2 \]

### Step 6: Solve for the spring constant \( k \)
Rearranging the equation to solve for \( k \):
\[ 5 = \frac{1}{2} k (0.25) \]
\[ 5 = 0.125 k \]
\[ k = \frac{5}{0.125} = 40 \text{ N/m} \]

### Final Answer
The force constant of the spring is:
\[ \boxed{40 \text{ N/m}} \]