A body of mass 5 kg is raised vertically to a height of 10 m by a force 170 N. The velocity of the body at this height will be

Options

A. 9.8 `ms^(-1)`
B. 15 `ms^(-1)`
C. 22 `ms^(-1)`
D. 37 `ms^(-1)`

Explanation

To solve the problem step by step, we will follow the principles of work, energy, and the forces acting on the body.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the body (m) = 5 kg
- Height (h) = 10 m
- Applied force (F) = 170 N
- Gravitational acceleration (g) = 10 m/s² (approximately)

2. **Calculate the Weight of the Body:**
The weight (W) of the body can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the values:
\[ W = 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 50 \, \text{N} \]

3. **Calculate the Net Force Acting on the Body:**
The net force (F_net) acting on the body when it is lifted can be calculated as:
\[ F_{\text{net}} = F - W \]
Substituting the values:
\[ F_{\text{net}} = 170 \, \text{N} - 50 \, \text{N} = 120 \, \text{N} \]

4. **Calculate the Work Done by the Net Force:**
The work done (W_net) by the net force in lifting the body to a height of 10 m is given by:
\[ W_{\text{net}} = F_{\text{net}} \cdot h \]
Substituting the values:
\[ W_{\text{net}} = 120 \, \text{N} \cdot 10 \, \text{m} = 1200 \, \text{J} \]

5. **Relate Work Done to Change in Kinetic Energy:**
According to the work-energy theorem, the work done on the body is equal to the change in kinetic energy (ΔK):
\[ W_{\text{net}} = \Delta K = K_f - K_i \]
Since the body starts from rest, the initial kinetic energy (K_i) is 0. Therefore:
\[ W_{\text{net}} = K_f \]
\[ 1200 \, \text{J} = \frac{1}{2} m v^2 \]

6. **Solve for the Final Velocity (v):**
Rearranging the equation to solve for v:
\[ 1200 = \frac{1}{2} \cdot 5 \cdot v^2 \]
\[ 1200 = \frac{5}{2} v^2 \]
\[ 2400 = 5 v^2 \]
\[ v^2 = \frac{2400}{5} = 480 \]
\[ v = \sqrt{480} \approx 22 \, \text{m/s} \]

### Final Answer:
The velocity of the body at a height of 10 m will be approximately **22 m/s**.