In the above problem if `m=1kg` and teminal velocity attained by its is 4m/s after falling a height of 1m, the energy dissipated as heat till then is `(g=10m//s^(2))`

Options

A. 10J
B. 2J
C. epsilonJ
D. 12J

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass (m) = 1 kg
- Terminal velocity (v) = 4 m/s
- Height (h) = 1 m
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the potential energy (PE) at the height of 1 meter
The potential energy (PE) at a height h is given by the formula:
\[ PE = mgh \]
Substituting the values:
\[ PE = 1 \, \text{kg} \times 10 \, \text{m/s}^2 \times 1 \, \text{m} = 10 \, \text{J} \]

### Step 3: Calculate the kinetic energy (KE) at terminal velocity
The kinetic energy (KE) when the object reaches terminal velocity is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 1 \, \text{kg} \times (4 \, \text{m/s})^2 = \frac{1}{2} \times 1 \times 16 = 8 \, \text{J} \]

### Step 4: Apply the work-energy theorem
According to the work-energy theorem, the work done by external forces is equal to the change in mechanical energy (kinetic + potential). In this case, the work done by resistive forces (like air resistance) will be equal to the initial potential energy minus the kinetic energy at terminal velocity:
\[ \text{Work done by resistive force} = KE + PE_{final} - PE_{initial} \]
Since the final potential energy (PE) at the terminal point is 0 (as it has fallen 1 meter):
\[ \text{Work done by resistive force} = 8 \, \text{J} + 0 - 10 \, \text{J} \]
\[ \text{Work done by resistive force} = -2 \, \text{J} \]

### Step 5: Determine the energy dissipated as heat
The negative sign indicates that energy is being dissipated (lost) due to resistive forces. Therefore, the energy dissipated as heat is:
\[ \text{Energy dissipated} = 2 \, \text{J} \]

### Final Answer:
The energy dissipated as heat till the object reaches terminal velocity after falling 1 meter is **2 Joules**.

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