Suppose a car approaches a hill and has an initial speed of 108 km/h at the bottom of the hill. The driver takes her foot off of
the gas pedal and allows the car to coast up the hill.a. If the car has the initial speed stated at a height of h = 0, how high, in meters, can the car coast up a hill if work done by friction is negligible?b. If, in actuality, a 710 kg car with an initial speed of 108 km/h is observed to coast up a hill and stops at a height 21 m above its starting point, how much thermal energy was generated by friction in J?c. What is the magnitude of the average force of friction, in newtons, if the hill has a slope 2.9 degrees above the horizontal?
1 answer:
Answer:
a) The car will reach a height of 45.9 m.
b) The amount of thermal energy generated is 173382 J.
c) The magnitude of the force of friction is 417.8 N.
Explanation:
Hi there!
a) In this problem, we have to use the conservation of energy. The energy conservation theorem states that the energy of a system remains constant. Energy can´t be created nor destroyed, only transformed. In the case of the car, the initial kinetic energy is transformed into potential energy as the car´s height increases while coasting up the hill.
Then, all the initial kinetic energy (KE) will be transformed into potential energy (PE) (only if there is no friction).
The equation of KE is the following:
KE = 1/2 · m · v²
Where:
m = mass of the car.
v = speed of the car.
The equation of PE is the following:
PE = m · g · h
Where:
m = mass of the car.
g = acceleration due to gravity.
h = height at which the car is located.
Since work done by friction is negligible, we can assume that all the initial kinetic energy will be transformed into potential energy. Then:
KE at the bottom of the hill = PE at the top of the hill
1/2 · m · v² = m · g · h
Solving for h:
1/2 · v² / g = h
Let´s convert the speed unit into m/s:
108 km/h · 1000 m/ 1 km · 1 h / 3600 s = 30 m/s
Now, let´s calculate h:
h = 1/2 · (30 m/s)² / 9.8 m/s²
h = 45.9 m
The car will reach a height of 45.9 m.
b) In this case, all the kinetic energy is not transformed into potential energy because some energy is transformed into thermal energy due to friction. The thermal energy generated is equal to the work done by friction. Then:
KE at the bottom of the hill = PE + work done by friction
KE = PE + Wfr (where Wfr is the work done by friction).
1/2 · m · v² = m · g · h + Wfr
1/2 · m · v² - m · g · h = Wfr
1/2 · 710 kg · (30 m/s)² - 710 kg · 9.8 m/s² · 21 m = Wfr
Wfr = 173382 J
The amount of thermal energy generated is 173382 J.
c) The work done by friction is calculated as follows:
Wfr = Ffr · Δx
Where:
Ffr = friction force.
Δx = traveled distance
Please, see the attached figure to notice that the traveled distance can be calculated by trigonometry using this trigonometric rule of right triangles:
sin angle = opposite side / hypotenuse
In our case:
sin 2.9° = h / Δx
Δx = h / sin 2.9°
Δx = 21 m / sin 2.9° = 415 m
Then, solving for the friction force using the equation of the work done by friction:
Wfr = Ffr · Δx
Wfr / Δx = Ffr
173382 J / 415 m = Ffr
Ffr = 417.8 N
The magnitude of the force of friction is 417.8 N
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Answer:
The ball will drop 0.881 m by the time it reaches the catcher.
Explanation:
The position of the ball at time "t" is described by the position vector "r":
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
When the ball reaches the catcher, the position vector will be "r final" (see attached figure).
The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.
Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):
x = x0 + v0x · t
17.0 m = 0 m + 40.1 m/s · t
t = 17.0 m/ 40. 1 m/s = 0.424 s
With this time, we can calculate the y-component of the vector "r final", the drop of the ball:
y = y0 + v0y · t + 1/2 · g · t²
Initially, there is no vertical velocity, then, v0y = 0.
y = 1/2 · g · t²
y = -1/2 · 9.8 m/s² · (0.424 s)²
y = -0.881 m
The ball will drop 0.881 m by the time it reaches the catcher.
