(a) How high a hill can a car coast up (engine disengaged) if friction is negligible and its initial speed is 78.0 km/h?
m
(b) If, in actuality, a 750 kg car with an initial speed of 78.0 km/h is observed to coast up a hill to a height 12.0 m above its starting point, how much thermal energy was generated by friction?
J
(c) What is the average force of friction if the hill has a slope 2.5° above the horizontal? (Explicitly show on paper how you follow the steps in the Problem-Solving Strategy for energy found on pages 159 and 160. Your instructor may ask you to turn in this work.)
(a) To determine the maximum height a car can coast up a hill with negligible friction, we need to analyze the energy involved.
1. Start by converting the car's initial speed from km/h to m/s:
Speed = 78.0 km/h × (1000 m/1 km) × (1 h/3600 s) = 21.7 m/s
2. The total mechanical energy (E) of the car is conserved during the coasting motion. It can be expressed as the sum of its kinetic energy (KE) and potential energy (PE):
E = KE + PE
3. The car's kinetic energy is given by:
KE = (1/2)mv^2
where m is the car's mass and v is the velocity.
Since the car is coasting with the engine disengaged, KE = 0.
4. The potential energy of the car due to its height above a reference point (usually the starting point) is given by:
PE = mgh
where m is the car's mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.
Rearrange this equation to solve for h:
h = PE /(mg)
5. Substitute the values into the equation:
h = (12.0 m) / [(750 kg) × (9.8 m/s^2)]
= 0.00172 m (rounded to four significant figures)
Therefore, the car can coast up a hill with a maximum height of approximately 0.00172 meters.
(b) To determine the thermal energy generated by friction, we can consider the energy lost by the car as it coasts up the hill.
1. Calculate the gravitational potential energy (PE) gained by the car:
PE = mgh
= (750 kg) × (9.8 m/s^2) × (12.0 m)
= 88,200 J
2. Total mechanical energy is conserved, so the amount of energy lost to thermal energy (TE) due to friction is equal to the difference between the initial mechanical energy and the final mechanical energy:
TE = Initial mechanical energy - Final mechanical energy
3. The initial mechanical energy is given by the initial kinetic energy (KE):
Initial mechanical energy = KE = (1/2)mv^2
4. Convert the initial speed from km/h to m/s:
Speed = 78.0 km/h × (1000 m/1 km) × (1 h/3600 s) = 21.7 m/s
5. Calculate the initial mechanical energy:
Initial mechanical energy = (1/2) × (750 kg) × (21.7 m/s)^2
= 214,583 J
6. The final mechanical energy is the sum of the gravitational potential energy (PE) gained and the thermal energy (TE) lost:
Final mechanical energy = PE + TE = 88,200 J + TE
7. Rearrange the equation to solve for TE:
TE = Final mechanical energy - PE
= 214,583 J - 88,200 J
= 126,383 J
Therefore, the thermal energy generated by friction is approximately 126,383 J.
(c) To determine the average force of friction, we can use the information given about the slope of the hill.
1. The gravitational force (Fg) acting on the car can be calculated using the formula:
Fg = mg
= (750 kg) × (9.8 m/s^2)
= 7350 N
2. Calculate the component of the gravitational force along the slope (F_parallel) using trigonometry:
F_parallel = Fg × sin(θ)
= 7350 N × sin(2.5°)
3. The work done by the friction force (W_friction) is equal to the gravitational force's parallel component multiplied by the distance traveled along the slope (d). In this case, we can assume d = h (the height gained):
W_friction = F_parallel × d
4. The work done by the friction force is also equal to the thermal energy generated (TE):
W_friction = TE
Therefore,
TE = F_parallel × h
Rearrange this equation to solve for F_parallel:
F_parallel = TE / h
5. Substitute the values into the equation:
F_parallel = (126,383 J) / (12.0 m)
= 10,532 N
Therefore, the average force of friction is approximately 10,532 N.
(a) To determine how high a hill a car can coast up with an engine disengaged and negligible friction, we can use the principle of conservation of energy.
The initial speed of the car is given as 78.0 km/h. To find the corresponding speed in m/s, we convert it:
speed (m/s) = (78.0 km/h) * (1000 m/km) * (1 h/3600 s)
= 21.7 m/s
Since we're assuming negligible friction, the only force acting on the car is gravity. The potential energy gained by the car as it climbs the hill is equal to the loss of its kinetic energy.
Let's consider the reference point at the bottom of the hill as the zero potential energy level. At the top of the hill, the car's kinetic energy becomes zero.
Using the equation for kinetic energy:
Initial kinetic energy = Final potential energy
(1/2) * m * v^2 = m * g * h
Where:
m = mass of the car
v = final velocity of the car (0 m/s at the top of the hill)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the hill
Rearranging the equation to solve for h:
h = (1/2) * v^2 / g
Substituting the known values:
h = (1/2) * (21.7 m/s)^2 / 9.8 m/s^2
h ≈ 24.1 m
Therefore, a car can coast up a hill approximately 24.1 meters high with an engine disengaged and negligible friction.
(b) To find the thermal energy generated by friction as the 750 kg car coasts up a hill to a height of 12.0 m above its starting point, we need to consider the loss of mechanical energy due to friction.
The total mechanical energy at the bottom of the hill is equal to the sum of the car's kinetic energy and gravitational potential energy:
Initial mechanical energy = (1/2) * m * v^2 + m * g * h
The final mechanical energy at the top of the hill is only gravitational potential energy:
Final mechanical energy = m * g * h_f
By conservation of energy, the thermal energy generated by friction is equal to the loss of mechanical energy:
Thermal energy = Initial mechanical energy - Final mechanical energy
Substituting the known values:
Thermal energy = (1/2) * (750 kg) * (21.7 m/s)^2 + (750 kg) * (9.8 m/s^2) * (12.0 m) - (750 kg) * (9.8 m/s^2) * (0 m)
Thermal energy ≈ 9.80 * 10^4 J
Therefore, approximately 9.80 * 10^4 Joules of thermal energy was generated by friction in this scenario.
(c) To find the average force of friction on a hill with a 2.5° slope, we can use the relationship between gravitational force and friction force on an incline.
The force of gravity acting on the car can be split into two components:
- The component parallel to the slope, which is responsible for accelerating the car up the hill.
- The component perpendicular to the slope, which is countered by the normal force and doesn't contribute to friction.
In this case, the force parallel to the slope can be calculated using the equation:
force_parallel = m * g * sin(θ)
Where:
m = mass of the car
g = acceleration due to gravity (9.8 m/s^2)
θ = angle of the slope in radians (2.5° converted to radians: 2.5° * π/180)
Substituting the known values:
force_parallel = (750 kg) * (9.8 m/s^2) * sin(2.5° * π/180)
Calculating the force parallel to the slope:
force_parallel ≈ 240 N
The force of friction is equal in magnitude but opposite in direction to the force parallel to the slope. Therefore, the average force of friction acting on the car is approximately 240 N, in the direction opposite to its motion up the hill.