Answer :
The tractor performed work of approximately 1.3875 x 10^12 Joules (or 1387.5 Terajoules) in dragging the 25-ton load across 3700 kilometers.
Work done (W) is calculated by the formula: W = F x d, where F is the force exerted and d is the distance traveled in the direction of the force.
Here's how we find the work done by the tractor:
1. Force Calculation:
- First, convert the mass (m) of the load from tons to kilograms: m = 25 tons 1000 kg/ton = 25,000 kg
- The force due to friction (F_f) opposes the motion and is given by: F_f = µ N, where µ is the friction coefficient (0.175) and N is the normal force acting on the load (equal to its weight).
- Since the tractor drags the load horizontally, the normal force is equal to the weight: N = m g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
Therefore, F_f = 0.175 * 25,000 kg 9.8 m/s² = 428750 N.
2. Distance Conversion:
- We need the distance (d) in meters for the work calculation. Convert kilometers to meters: d = 3700 km * 1000 m/km = 3,700,000 m.
3. Work Calculation:
- Now that we have both force and distance, apply the work formula: W = F_f d = 428750 N * 3,700,000 m = 1.59375 x 10^12 J
Since Joules are very large units for this scenario, it's common practice to express the answer in Terajoules (TJ). One Terajoule is equal to 10^12 Joules. Therefore, the work done by the tractor is approximately 1.3875 x 10^12 Joules (or 1387.5 Terajoules).
The tractor did [tex]\( 158.92 \times 10^9 \)[/tex] Joules of work.
To calculate the work done by the tractor, we use the formula:
[tex]\[ \text{Work} = \text{Force} \times \text{Distance} \][/tex]
The force of friction can be calculated using:
[tex]\[ \text{Force} = \text{Friction coefficient} \times \text{Normal force} \][/tex]
Given the mass of the sled is [tex]\( 25 \times 1000 \, \text{kg} = 25000 \, \text{kg} \), and \( g = 9.81 \, \text{m/s}^2 \)[/tex]:
[tex]\[ \text{Normal force} = \text{mass} \times g \]\[ = 25000 \times 9.81 \]\[ = 245250 \, \text{N} \]\[ \text{Force} = 0.175 \times 245250 \]\[ = 42937.5 \, \text{N} \][/tex]
Now, converting distance to meters:
[tex]\[ \text{Distance} = 3700 \times 1000 \]\[ = 3700000 \, \text{m} \][/tex]
Now, calculate work:
[tex]\[ \text{Work} = 42937.5 \times 3700000 \]\[ = 158918750000 \, \text{J} \][/tex]
So, the tractor did [tex]\( 158.92 \times 10^9 \)[/tex] Joules of work.
