1000 N force is required to lift a hook nd 10000 N force is reuires to lift a load slowly.Find power required to lift hook with load with speed v=0.5`m//s`

Options

A. 5kW
B. 1.5 kW
C. 5.5 kW
D. 4.5 kW

Answer

To solve the problem, we need to calculate the power required to lift a hook and a load at a given speed. Let's break this down step by step.

### Step 1: Identify the forces involved
- The force required to lift the hook is \( F_{\text{hook}} = 1000 \, \text{N} \).
- The force required to lift the load is \( F_{\text{load}} = 10000 \, \text{N} \).

### Step 2: Calculate the total force
The total force required to lift both the hook and the load is the sum of the individual forces:
\[ F_{\text{total}} = F_{\text{hook}} + F_{\text{load}} = 1000 \, \text{N} + 10000 \, \text{N} = 11000 \, \text{N} \]

### Step 3: Determine the speed
The speed at which the hook and load are lifted is given as:
\[ v = 0.5 \, \text{m/s} \]

### Step 4: Calculate the power
Power is defined as the product of force and velocity. The formula for power \( P \) is:
\[ P = F \cdot v \]
Substituting the values we have:
\[ P = F_{\text{total}} \cdot v = 11000 \, \text{N} \cdot 0.5 \, \text{m/s} \]
Calculating this gives:
\[ P = 5500 \, \text{W} \]

### Step 5: Convert power to kilowatts
Since power is often expressed in kilowatts (kW), we convert watts to kilowatts:
\[ P = 5500 \, \text{W} = 5.5 \, \text{kW} \]

### Final Answer
The power required to lift the hook with the load at a speed of \( 0.5 \, \text{m/s} \) is:
\[ \boxed{5.5 \, \text{kW}} \]