The 2500 kg cable car shown in the figure below descends a 200 m high hill. In addition to its brakes, the cable car controls its speed by pulling an 1200 kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.
One day the brakes fail just as the cable car leaves the top on its downward journey. What is the runaway car's speed at the bottom of the hill?
Well, looks like the cable car is taking a joyride! Let's calculate its runaway speed, shall we?
First, we can use the principle of conservation of energy to solve this problem. The potential energy at the top of the hill is converted into kinetic energy at the bottom. We'll ignore any energy losses due to friction or air resistance because, you know, the cable car doesn't believe in friction.
The potential energy of the cable car at the top of the hill is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. So, the potential energy of the cable car is (2500 kg) * (9.8 m/s²) * (200 m).
This energy is then converted into kinetic energy at the bottom. The kinetic energy is given by (1/2)mv², where v is the velocity of the cable car at the bottom.
Setting the potential energy equal to the kinetic energy, we get:
(2500 kg) * (9.8 m/s²) * (200 m) = (1/2)(2500 kg)v².
Now, let's solve for v:
v² = (2 * (2500 kg) * (9.8 m/s²) * (200 m)) / (2500 kg)
v² = 2 * 9.8 m/s² * 200 m
v² = 2 * 9.8 * 200 m²/s²
v² = 3920 m²/s²
Taking the square root of both sides, we find:
v = √(3920 m²/s²)
v ≈ 62.6 m/s
So, the runaway cable car will be zooming down the hill at approximately 62.6 m/s at the bottom. Watch out below!
To find the runaway car's speed at the bottom of the hill, we can use the principle of conservation of energy. The potential energy at the top of the hill is converted into kinetic energy at the bottom.
First, let's find the potential energy at the top of the hill. The potential energy (PE) is given by the formula:
PE = mgh
where m is the mass of the cable car and h is the height of the hill. In this case, the mass of the cable car is 2500 kg and the height of the hill is 200 m. Substituting in these values, we get:
PE = 2500 kg * 9.8 m/s^2 * 200 m = 4,900,000 Joules
Next, let's find the kinetic energy at the bottom of the hill. The kinetic energy (KE) is given by the formula:
KE = 1/2 * mv^2
where m is the mass of the cable car and v is its velocity at the bottom of the hill. In this case, the mass of the cable car is 2500 kg. We need to solve for v.
Using the conservation of energy principle, we can equate the potential energy at the top to the kinetic energy at the bottom:
PE = KE
4,900,000 J = 1/2 * 2500 kg * v^2
Now we can solve for v. Rearranging the equation:
v^2 = (4,900,000 J) / (1/2 * 2500 kg)
v^2 = 3920 m^2/s^2
v = √(3920 m^2/s^2) = 62.7 m/s (rounded to one decimal place)
Therefore, the runaway car's speed at the bottom of the hill is approximately 62.7 m/s.