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Sagot :
Sure, let's break down the steps to solve the problem step-by-step.
1. Identify the given data:
- Mass of the speedboat, [tex]\( m = 1.00 \times 10^3 \)[/tex] kg.
- Final velocity, [tex]\( v_f = 20.0 \)[/tex] m/s.
- Time taken, [tex]\( t = 5.00 \)[/tex] s.
- Constant drag force, [tex]\( F_d = 5.00 \times 10^2 \)[/tex] N.
2. Calculate the acceleration:
Since the speedboat goes from rest to 20.0 m/s in 5.00 seconds, we can use the following formula for constant acceleration:
[tex]\[ a = \frac{v_f - v_i}{t} \][/tex]
Here, the initial velocity [tex]\( v_i = 0 \)[/tex] (the boat starts from rest).
[tex]\[ a = \frac{20.0 \text{ m/s}}{5.00 \text{ s}} = 4.0 \text{ m/s}^2 \][/tex]
3. Calculate the net force needed to achieve this acceleration:
Using Newton's second law, [tex]\( F = ma \)[/tex]:
[tex]\[ F_{\text{net}} = m \cdot a = (1.00 \times 10^3 \text{ kg}) \times (4.0 \text{ m/s}^2) = 4000 \text{ N} \][/tex]
4. Calculate the total force needed to overcome both the drag and achieve the acceleration:
The total force [tex]\( F_{\text{total}} \)[/tex] needed is the sum of the net force [tex]\( F_{\text{net}} \)[/tex] and the drag force [tex]\( F_d \)[/tex]:
[tex]\[ F_{\text{total}} = F_{\text{net}} + F_d = 4000 \text{ N} + 500 \text{ N} = 4500 \text{ N} \][/tex]
5. Calculate the power needed:
Power [tex]\( P \)[/tex] is given by the formula:
[tex]\[ P = F_{\text{total}} \cdot v_f \][/tex]
[tex]\[ P = 4500 \text{ N} \times 20.0 \text{ m/s} = 90000 \text{ W} = 90.0 \text{ kW} \][/tex]
So, the power that the speedboat needs to go from rest to 20.0 m/s in 5.00 seconds, overcoming a constant drag force, is [tex]\(\boxed{90.0 \text{ kW}}\)[/tex].
1. Identify the given data:
- Mass of the speedboat, [tex]\( m = 1.00 \times 10^3 \)[/tex] kg.
- Final velocity, [tex]\( v_f = 20.0 \)[/tex] m/s.
- Time taken, [tex]\( t = 5.00 \)[/tex] s.
- Constant drag force, [tex]\( F_d = 5.00 \times 10^2 \)[/tex] N.
2. Calculate the acceleration:
Since the speedboat goes from rest to 20.0 m/s in 5.00 seconds, we can use the following formula for constant acceleration:
[tex]\[ a = \frac{v_f - v_i}{t} \][/tex]
Here, the initial velocity [tex]\( v_i = 0 \)[/tex] (the boat starts from rest).
[tex]\[ a = \frac{20.0 \text{ m/s}}{5.00 \text{ s}} = 4.0 \text{ m/s}^2 \][/tex]
3. Calculate the net force needed to achieve this acceleration:
Using Newton's second law, [tex]\( F = ma \)[/tex]:
[tex]\[ F_{\text{net}} = m \cdot a = (1.00 \times 10^3 \text{ kg}) \times (4.0 \text{ m/s}^2) = 4000 \text{ N} \][/tex]
4. Calculate the total force needed to overcome both the drag and achieve the acceleration:
The total force [tex]\( F_{\text{total}} \)[/tex] needed is the sum of the net force [tex]\( F_{\text{net}} \)[/tex] and the drag force [tex]\( F_d \)[/tex]:
[tex]\[ F_{\text{total}} = F_{\text{net}} + F_d = 4000 \text{ N} + 500 \text{ N} = 4500 \text{ N} \][/tex]
5. Calculate the power needed:
Power [tex]\( P \)[/tex] is given by the formula:
[tex]\[ P = F_{\text{total}} \cdot v_f \][/tex]
[tex]\[ P = 4500 \text{ N} \times 20.0 \text{ m/s} = 90000 \text{ W} = 90.0 \text{ kW} \][/tex]
So, the power that the speedboat needs to go from rest to 20.0 m/s in 5.00 seconds, overcoming a constant drag force, is [tex]\(\boxed{90.0 \text{ kW}}\)[/tex].
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