A cyclist starts from rest and coasts down a 6.5∘{\rm ^\circ} hill. The mass of the cyclist plus bicycle is 95kg{\rm kg} . The cyclist has traveled 270m{\rm m} .What was the net work done by gravity on the cyclist? How fast is the cyclist going? Ignore air resistance.
a=gsin(theta) not gcos(theta)
Note:
the degree symbol is written as
& d e g ;
(suppress all spaces)
Difference in elevation, Δh
= 270m × sin(6.5°)
total mass (cyclist + bicycle), m
= 95 kg
total weight
= mg
Work done (on cyclist & bicycle)
=mgΔh
Component of force along slope, F
= mg cos(θ)
downward acceleration along slope, a
= F/m
= g cos(θ)
Consider the acceleration over distance d of 270 m
d=270 m
vi=0 (initial velocity)
vf (final velocity)
a=gcos(θ)
θ=6.5°
g=9.8 m/s²
m=95 kg
We can apply the kinematics equation
vf²=vi²+2ad
to solve for vf.
To find the net work done by gravity on the cyclist, we need to calculate the change in potential energy of the cyclist as they go down the hill. This can be calculated using the formula:
ΔPE = mgh
Where:
ΔPE is the change in potential energy
m is the mass of the cyclist plus bicycle
g is the acceleration due to gravity
h is the height of the hill
Given:
m = 95 kg
g = 9.8 m/s^2
h = (270 m) * sin(6.5°) (since the height of the hill is the vertical distance traveled)
First, we calculate the height of the hill:
h = 270 m * sin(6.5°)
h ≈ 29.51 m
Substituting the values into the formula:
ΔPE = (95 kg) * (9.8 m/s^2) * (29.51 m)
ΔPE ≈ 27644.36 J
The net work done by gravity on the cyclist is equal to the change in potential energy, so the net work done is approximately 27644.36 Joules.
To find the speed of the cyclist at the bottom of the hill, we can use the principle of conservation of energy. The potential energy lost by the cyclist is converted into kinetic energy.
The kinetic energy (KE) is given by:
KE = 0.5mv^2
Where:
KE is the kinetic energy
m is the mass of the cyclist plus bicycle
v is the velocity of the cyclist
Equating the change in potential energy to the kinetic energy:
ΔPE = KE
mgh = 0.5mv^2
Simplifying the equation:
gh = 0.5v^2
Substituting the values:
(9.8 m/s^2) * (29.51 m) = 0.5v^2
Solving for v:
v^2 = (2 * 9.8 m/s^2 * 29.51 m)
v ≈ √(579.638 N m / 95 kg)
v ≈ √6.097 m^2/s^2
v ≈ 2.47 m/s
Therefore, the cyclist is going approximately 2.47 m/s at the bottom of the hill.
To find the net work done by gravity on the cyclist, we need to calculate the change in gravitational potential energy. The work done by gravity is equal to the change in potential energy.
The potential energy of an object at a certain height is given by the formula:
Potential Energy = mass × gravity × height
In this case, the height of the hill is not provided, but we can use some trigonometry to calculate it. Since we know the angle of the hill, we can use the formula:
Height = distance × sin(angle)
Plugging in the values, we get:
Height = 270m × sin(6.5∘)
Now, we can substitute the calculated height back into the potential energy formula:
Potential Energy = mass × gravity × height
Next, we need to calculate the final velocity of the cyclist. The initial velocity is given as zero since the cyclist starts from rest. We can use the formula for the final velocity of an object undergoing uniform acceleration:
Final velocity^2 = Initial velocity^2 + 2 × acceleration × distance
In this case, the acceleration is due to gravity, and the distance traveled is given as 270m. Since the cyclist is coasting downhill, we can assume that there's no external force acting on the cyclist, so the acceleration is equal to the acceleration due to gravity, which is approximately 9.8m/s^2.
By rearranging the formula, we can solve for the final velocity:
Final velocity^2 = 0 + 2 × 9.8m/s^2 × 270m
Now, we can take the square root of both sides to get the final velocity.
Once we have the final velocity, we can calculate the net work done by gravity using the work-energy principle:
Net Work = (1/2) × mass × final velocity^2
Now, we have the method to calculate both the net work done by gravity and the cyclist's speed. Let's calculate the values:
1. Calculate the height of the hill:
Height = 270m × sin(6.5∘)
2. Calculate the potential energy:
Potential Energy = mass × gravity × height
3. Calculate the final velocity:
Final velocity^2 = 0 + 2 × 9.8m/s^2 × 270m
Final velocity = √(Final velocity^2)
4. Calculate the net work done by gravity:
Net Work = (1/2) × mass × final velocity^2
You can plug in the values into the respective formulas to find the answers.