A rocket driven sledge speeds up from 40 metres per second to 55 metres per second in 5.0 seconds, using an engine that produces 3500 newtons of thrust. How much thrust would be needed to get the same increase in speed in 2.0 seconds

Options

A. 8550
B. 8750
C. 8700
D. 8500

Explanation

To solve the problem, we need to determine how much thrust (F2) is required to achieve the same change in speed of the rocket-driven sledge from 40 m/s to 55 m/s in 2 seconds, given that it takes 5 seconds with a thrust of 3500 N.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial speed (u) = 40 m/s
- Final speed (v) = 55 m/s
- Time for the first case (Δt1) = 5.0 seconds
- Thrust in the first case (F1) = 3500 N
- Time for the second case (Δt2) = 2.0 seconds

2. **Calculate the Change in Speed:**
- Change in speed (Δv) = Final speed - Initial speed
- Δv = 55 m/s - 40 m/s = 15 m/s

3. **Calculate the Change in Momentum:**
- The change in momentum (ΔP) can be expressed as:
\[ \Delta P = m \cdot \Delta v \]
- Since the mass (m) of the sledge is constant, we can relate the change in momentum in both cases.

4. **Relate the Change in Momentum to Thrust:**
- In the first case:
\[ \Delta P_1 = F_1 \cdot \Delta t_1 \]
- In the second case:
\[ \Delta P_2 = F_2 \cdot \Delta t_2 \]
- Since the change in momentum is the same in both cases:
\[ \Delta P_1 = \Delta P_2 \]

5. **Set Up the Equation:**
- From the above relationships:
\[ F_1 \cdot \Delta t_1 = F_2 \cdot \Delta t_2 \]
- Substituting the known values:
\[ 3500 \, \text{N} \cdot 5.0 \, \text{s} = F_2 \cdot 2.0 \, \text{s} \]

6. **Solve for F2:**
- Rearranging the equation to solve for F2:
\[ F_2 = \frac{3500 \, \text{N} \cdot 5.0 \, \text{s}}{2.0 \, \text{s}} \]
- Calculate:
\[ F_2 = \frac{17500 \, \text{N} \cdot \text{s}}{2.0 \, \text{s}} = 8750 \, \text{N} \]

### Final Answer:
The thrust needed to achieve the same increase in speed in 2.0 seconds is **8750 N**.