A 1400 kilogram car crests a 3200 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and cresting a 2800 m pass. (a) neglecting friction, what should the speed of the car be at the top of the second pass? (b) find the actual speed of the car if the work due to non-conservative forces is -5×10^6 J.
To solve this problem, we'll use the principle of conservation of mechanical energy. Mechanical energy is the sum of kinetic energy (KE) and potential energy (PE). Neglecting friction, the mechanical energy of the car will be conserved throughout the entire process.
Let's break the problem into two parts:
Part 1: Climbing the First Pass
The car starts from rest, so its initial kinetic energy is zero. The only form of energy present is potential energy.
PE1 = mgh1
where m is the mass of the car (1400 kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h1 is the height of the first pass (3200 m). Therefore, the potential energy at the top of the first pass is:
PE1 = (1400 kg)(9.8 m/s²)(3200 m) = 43,648,000 J
Part 2: Descending and Climbing the Second Pass
The car descends 1000 m, losing potential energy but gaining kinetic energy. At the bottom of the descent, all the potential energy is converted into kinetic energy:
KE2 = PE1 - PE_loss
where PE_loss is the loss of potential energy during the descent. Since the car comes to rest, the loss of potential energy is equal to the initial potential energy at the top of the first pass.
PE_loss = 43,648,000 J
Therefore, the kinetic energy at the bottom of the descent is:
KE2 = 43,648,000 J - 43,648,000 J = 0 J
Now, the car climbs the second pass. All the kinetic energy is converted back into potential energy at the top of the second pass:
PE2 = KE2 + PE_loss
Since KE2 is zero, the potential energy at the top of the second pass is:
PE2 = 43,648,000 J
(a) Neglecting Friction: Speed at the Top of the Second Pass
To find the speed at the top of the second pass, we need to use the equation for kinetic energy:
KE = 1/2mv²
Since the potential energy at the top of the second pass is the same as the initial potential energy, we can use the equation for potential energy to solve for v:
PE = mgh
Rearranging the equation, we get:
v = sqrt(2gh)
where g is the acceleration due to gravity (approximately 9.8 m/s²) and h is the height of the second pass (2800 m). Plugging in the values, we have:
v = sqrt(2 * 9.8 m/s² * 2800 m) ≈ 235.12 m/s
Therefore, neglecting friction, the speed of the car at the top of the second pass is approximately 235.12 m/s.
(b) Actual Speed at the Top of the Second Pass
To find the actual speed of the car at the top of the second pass, we need to consider the work done by non-conservative forces, such as friction. The work done by these forces is given as -5×10^6 J.
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy:
Work = ΔKE
Knowing that the kinetic energy at the bottom of the descent is zero, we can calculate the change in kinetic energy:
ΔKE = KE2 - KE1
Since KE2 is zero, we have:
ΔKE = -KE1
Therefore:
-5×10^6 J = -43,648,000 J
Solving for KE1:
KE1 = 43,648,000 J
Now, we can use KE1 to find the actual speed at the top of the second pass:
KE1 = 1/2mv²
Plugging in the values, we have:
43,648,000 J = 1/2 * 1400 kg * v²
Simplifying the equation, we get:
v² = (2 * 43,648,000 J) / (1400 kg)
v ≈ sqrt(62,354.29 m²/s²) ≈ 249.71 m/s
Therefore, taking into account the work done by non-conservative forces, the actual speed of the car at the top of the second pass is approximately 249.71 m/s.