In Greek Mythology, Sisyphus was a man who tricked death several times and was punished in the Underworld by being forced to eternally roll a boulder up a hill.

In Greek Mythology, Sisyphus was a man who tricked death several times and was punished in the Underworld by being forced to eternally roll a boulder up a hill. But this was an enchanted boulder so whenever he was close to reaching the top, the boulder would roll away from him back down the hill. For this problem, we are going to model this punishment. We are going to ignore the rolling and model this as a sliding boulder. The hill has a constant slope inclined at 30 degrees above the horizontal and the mass of the boulder is 839 kg. While Sisyphus is pushing the boulder up the incline, he exerts a constant force of 5855 N up the incline and the boulder moves up the incline with constant velocity. Assume the acceleration due to gravity in the Underworld is the same as it is on the surface of the earth. Calculate the coefficient of kinetic friction between boulder and the hill. Tries 2/10 Previous Tries Once the boulder gets away from Sisyphus, calculate the magnitude of the acceleration of the boulder down the hill. Tries 0/10 Let’s assume that once the boulder leaves Sisyphus that it starts from rest. If the boulder then travels 148 m down the hill, calculate the time it takes the boulder to travel this distance. Tries 0/10

The Correct Answer and Explanation is:

We are given a situation where a boulder is sliding up an incline with a constant velocity, and we are asked to calculate the coefficient of kinetic friction, the acceleration of the boulder as it slides down, and the time it takes the boulder to travel 148 m down the hill after being released. Here’s how we can solve this:

Step 1: Calculate the coefficient of kinetic friction (μk)

The boulder moves up the incline at a constant velocity, which implies that the net force acting on it is zero (since there’s no acceleration). The forces involved include:

1. The gravitational force: This has a component along the incline, ( F_{\text{gravity, parallel}} = mg \sin(\theta) ), where ( m ) is the mass of the boulder (839 kg), ( g ) is the acceleration due to gravity (9.8 m/s²), and ( \theta = 30^\circ ) is the angle of the incline.
2. The frictional force: The friction opposes the motion of the boulder and is given by ( F_{\text{friction}} = \mu_k N ), where ( N ) is the normal force, and ( \mu_k ) is the coefficient of kinetic friction. The normal force is given by ( N = mg \cos(\theta) ).
3. The applied force: Sisyphus applies a constant force of 5855 N up the incline.

Since the boulder is moving with constant velocity, the forces along the incline must balance. Therefore, the sum of forces in the direction of motion must be zero:

[
F_{\text{applied}} = F_{\text{gravity, parallel}} + F_{\text{friction}}
]

Substitute the expressions for the forces:

[
5855 \, \text{N} = mg \sin(\theta) + \mu_k mg \cos(\theta)
]

Now, plug in the values for mass (( m = 839 \, \text{kg} )), ( g = 9.8 \, \text{m/s}^2 ), and ( \theta = 30^\circ ):

[
5855 = (839 \times 9.8 \times \sin(30^\circ)) + \mu_k (839 \times 9.8 \times \cos(30^\circ))
]

Calculate the gravitational components:

[
F_{\text{gravity, parallel}} = 839 \times 9.8 \times 0.5 = 4111.1 \, \text{N}
]
[
F_{\text{normal}} = 839 \times 9.8 \times 0.866 = 7159.5 \, \text{N}
]

Now, substitute these values into the equation:

[
5855 = 4111.1 + \mu_k \times 7159.5
]

Solve for ( \mu_k ):

[
5855 – 4111.1 = \mu_k \times 7159.5
]
[
1743.9 = \mu_k \times 7159.5
]
[
\mu_k = \frac{1743.9}{7159.5} \approx 0.243
]

So, the coefficient of kinetic friction ( \mu_k ) is approximately 0.243.

Step 2: Calculate the acceleration of the boulder down the hill

When the boulder is released and begins to slide down the hill, the forces acting on it are:

1. Gravitational force down the incline: ( F_{\text{gravity, parallel}} = mg \sin(\theta) ).
2. Frictional force opposing the motion: ( F_{\text{friction}} = \mu_k N = \mu_k mg \cos(\theta) ).

The net force acting on the boulder is:

[
F_{\text{net}} = mg \sin(\theta) – \mu_k mg \cos(\theta)
]

Using ( \mu_k = 0.243 ) and the same values for ( m ), ( g ), and ( \theta ):

[
F_{\text{net}} = (839 \times 9.8 \times \sin(30^\circ)) – (0.243 \times 839 \times 9.8 \times \cos(30^\circ))
]
[
F_{\text{net}} = 4111.1 – 0.243 \times 7159.5
]
[
F_{\text{net}} = 4111.1 – 1743.9 = 2367.2 \, \text{N}
]

Now, use Newton’s second law to find the acceleration:

[
F_{\text{net}} = ma
]

[
a = \frac{F_{\text{net}}}{m} = \frac{2367.2}{839} \approx 2.82 \, \text{m/s}^2
]

So, the acceleration of the boulder down the hill is approximately 2.82 m/s².

Step 3: Calculate the time to travel 148 m down the hill

To find the time it takes for the boulder to travel 148 m, we use the kinematic equation:

[
d = \frac{1}{2} a t^2
]

Solving for ( t ):

[
t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2 \times 148}{2.82}} \approx \sqrt{\frac{296}{2.82}} \approx \sqrt{104.93} \approx 10.24 \, \text{seconds}
]

So, the time it takes for the boulder to travel 148 m down the hill is approximately 10.24 seconds.

Summary of Results:

• The coefficient of kinetic friction between the boulder and the hill is approximately 0.243.
• The acceleration of the boulder down the hill is approximately 2.82 m/s².
• The time it takes for the boulder to travel 148 m down the hill is approximately 10.24 seconds.