Two railroad cars, each of mass 2.2 104 kg, are connected by a cable to each other, and the car in front is connected by a cable to the engine as shown in the figure below. The cars start from rest and accelerate to a speed of 1.7 m/s after 1 min.
a)Find the work done by cable 1 on the car in the back. ____J
b)Find the work done by cable 1 on the car in front. _____J
c)Find the work done by cable 2 on the car in the front. _____J
To find the work done by the cables on the railroad cars, we use the formula:
Work = Force × Distance × cos(θ)
where
Work is the work done,
Force is the net force applied,
Distance is the displacement of the object in the direction of the force, and
θ is the angle between the force applied and the direction of displacement.
Let's analyze each case separately:
a) Work done by cable 1 on the car in the back:
First, let's find the net force applied to the car in the back. The net force is equal to the mass of the car multiplied by its acceleration:
F_net = m × a
= (2.2 × 10^4 kg) × (1.7 m/s^2)
= 3.74 × 10^4 N
Since cable 1 connects the car in the back to the engine, the direction of the force applied by cable 1 is in the direction of motion of the car. Therefore, θ = 0 degrees.
The car moves a certain distance during the acceleration period. Assuming it moves in a straight line, the distance traveled is given by:
Distance = (initial velocity × time) + (0.5 × acceleration × time^2)
= (0 × 60 s) + (0.5 × 1.7 m/s^2 × (60 s)^2)
= 3060 m
Now we can calculate the work done by cable 1 on the car in the back:
Work = Force × Distance × cos(θ)
= (3.74 × 10^4 N) × (3060 m) × cos(0 degrees)
= 1.139 × 10^8 J
Hence, the work done by cable 1 on the car in the back is 1.139 × 10^8 Joules (J).
b) Work done by cable 1 on the car in the front:
The situation is the same as in part a), as cable 1 connects the car in the back to the engine. Therefore, the work done by cable 1 on the car in the front is also 1.139 × 10^8 J.
c) Work done by cable 2 on the car in the front:
Since cable 2 connects the two cars, it pulls the car in the front backward. The net force applied by cable 2 can be calculated similarly to part a):
F_net = m × a
= (2.2 × 10^4 kg) × (1.7 m/s^2)
= 3.74 × 10^4 N
In this case, the car moves in the opposite direction to the force applied by cable 2, so θ = 180 degrees.
Again, the distance traveled is 3060 m.
Now we can calculate the work done by cable 2 on the car in the front:
Work = Force × Distance × cos(θ)
= (3.74 × 10^4 N) × (3060 m) × cos(180 degrees)
= -1.139 × 10^8 J
Since cos(180 degrees) = -1, the work done by cable 2 on the car in the front is -1.139 × 10^8 J.
Hence, the work done by cable 2 on the car in the front is -1.139 × 10^8 Joules (J).