Certainly! Let's break down the problem step-by-step.
Firstly, let's state the given information:
1. Mass of the truck, [tex]\( m = 1.20 \times 10^3 \, \text{kg} \)[/tex]
2. Force applied, [tex]\( F = 25.2 \, \text{N} \)[/tex]
3. Maximum speed, [tex]\( v_{\text{max}} = 60.0 \, \text{km/h} \)[/tex]
### Part (i): Calculating Work Done by the Force
The work done by a force when it accelerates an object can be calculated using the work-energy theorem. This theorem states that the work done on an object is equal to the change in its kinetic energy.
Since the truck starts from rest, its initial kinetic energy is zero. The formula for the kinetic energy ([tex]\( K \)[/tex]) is given by:
[tex]\[ K = \frac{1}{2} m v^2 \][/tex]
We need to convert the maximum speed from km/h to m/s. The conversion is as follows:
[tex]\[ v_{\text{max\_ms}} = \frac{60.0 \, \text{km/h} \times 1000 \, \text{m/km}}{3600 \, \text{s/h}} = 16.6667 \, \text{m/s} \][/tex]
Now, substituting the values into the kinetic energy formula:
[tex]\[ K = \frac{1}{2} \times 1.20 \times 10^3 \, \text{kg} \times (16.6667 \, \text{m/s})^2 \][/tex]
[tex]\[ K = \frac{1}{2} \times 1.20 \times 10^3 \times 277.7778 \][/tex]
[tex]\[ K \approx 166666.67 \, \text{J} \][/tex]
Therefore, the work done by the force is approximately:
[tex]\[ W \approx 166666.67 \, \text{J} \][/tex]
### Part (ii): Calculating the Distance Travelled
To find the distance travelled by the truck to reach the maximum speed, we use the kinematic equation:
[tex]\[ v^2 = u^2 + 2 a d \][/tex]
Where:
- [tex]\( v = v_{\text{max\_ms}} \)[/tex] (final velocity)
- [tex]\( u = 0 \, \text{m/s} \)[/tex] (initial velocity)
- [tex]\( a \)[/tex] is acceleration
- [tex]\( d \)[/tex] is distance
First, let's find the acceleration. Using Newton's second law, [tex]\( F = m a \)[/tex]:
[tex]\[ a = \frac{F}{m} = \frac{25.2 \, \text{N}}{1.20 \times 10^3 \, \text{kg}} \][/tex]
[tex]\[ a \approx 0.021 \, \text{m/s}^2 \][/tex]
Substitute the values into the kinematic equation:
[tex]\[ (16.6667 \, \text{m/s})^2 = 0 + 2 \times 0.021 \, \text{m/s}^2 \times d \][/tex]
[tex]\[ 277.7778 = 0.042 d \][/tex]
[tex]\[ d = \frac{277.7778}{0.042} \][/tex]
[tex]\[ d \approx 6613.76 \, \text{m} \][/tex]
Therefore, the distance travelled by the truck to reach the maximum speed is approximately:
[tex]\[ d \approx 6613.76 \, \text{m} \][/tex]
### Summary
- The work done by the force is approximately [tex]\( 166666.67 \, \text{J} \)[/tex].
- The distance travelled by the truck in reaching the maximum speed is approximately [tex]\( 6613.76 \, \text{m} \)[/tex].
