Answer:
Energy Lost for group A's car = 0.687 J
Energy Lost for group B's car = 0.55 J
Explanation:
The exact question is as follows :
Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.
To find - How much energy is lost due to heat for group A's car ?
How much for Group B's car ?
Solution -
We know that,
GPE = 1 Joule (Potential Energy)
Now,
For Group A -
Energy Lost = GPE - KE
= 1 J - 0.313 J
= 0.687 J
So,
Energy Lost for group A's car = 0.687 J
Now,
For Group B -
Energy Lost = GPE - KE
= 1 J - 0.45 J
= 0.55 J
So,
Energy Lost for group B's car = 0.55 J
Most reactions are exothemic. If the forward reaction of an equilibrium reaction is exothemic then the reverse reaction must be endothermic.
If a system in equilibrium is heated, it will move in exothermic direction to give out heat energy.
1. Find the force of friction between the sports car and the station wagon stuck together and the road. The total mass m = 1928kg + 1041kg = 2969kg. The only force in the x-direction is friction: F = μ*N = μ * m * g
2. Find the acceleration due to friction:
F = m*a = μ * m * g => a = μ * g = 0.6 * 9.81
3. Find the time it took the two cars stuck together to slide 12m:
x = 0.5*a*t²
t = sqrt(2*x / a) = sqrt(2 * x / (μ * g) )
4. Find the initial velocity of the two cars:
v = a*t = μ * g * sqrt(2 * x / (μ * g) ) = sqrt( 2 * x * μ * g)
5. Use the initial velocity of the two cars combined to find the velocity of the sports car. Momentum must be conserved:
m₁ mass of sports car
v₁ velocity of sports car before the crash
m₂ mass of station wagon
v₂ velocity of station wagon before the crash = 0
v velocity after the crash
m₁*v₁ + m₂*v₂ = (m₁+m₂) * v = m₁*v₁
v₁ = (m₁+m₂) * v / m₁ = (m₁+m₂) * sqrt( 2 * x * μ * g) / m₁
v₁ = 33.9 m/s
Answer:
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