Question

Please help me with the steps to this problem.

A 80-kg skateboarder grinds down a hubba ledge that is 2.2 m long and inclined at 22 ∘ below the horizontal. Kinetic friction dissipates half of her initial potential energy to thermal and sound energies.

What is the coefficient of kinetic friction between her skateboard and the ledge surface?

Express your answer to three significant figures.

Answer 1

Answer 2

Given:

• Mass of skateboarder ($m$): 80 kg

• Length of ledge ($L$): 2.2 m

• Incline angle ($\theta$): 22° below the horizontal

• Energy dissipation: Half of initial potential energy ($U_i$) is lost to friction.

• 
  

---

Step 1: Calculate Initial Potential Energy ($U_i$)

First, determine the vertical height ($h$) of the ledge using trigonometry:

$$h=L\sin \theta =2.2\sin (22\mathrm{°})=2.2\times 0.3746\approx 0.824\text{m}$$

Now, compute $U_i$:

$$U_i=mgh=80\times 9.81\times 0.824\approx 647.5\text{J}$$

---

Step 2: Determine Energy Lost to Friction ($W_f$)

Half of $U_i$ is dissipated:

$$W_f=\frac{1}{2}{U}_i=\frac{647.5}{2}=323.75\text{J}$$

---

Step 3: Relate Friction Work to Frictional Force

The work done by friction is:

$$W_f=f_k\times L$$

Where $f_k$ is the kinetic friction force:

$$f_k={\mu }_{k}N$$

Normal force ($N$) balances the perpendicular component of weight:

$$N=mg\cos \theta =80\times 9.81\times \cos (22\mathrm{°})\approx 728.5\text{N}$$

Substitute $W_f$ and solve for ${\mu }_k$:

$$323.75={\mu }_k\times 728.5\times 2.2$$$${\mu }_k=\frac{323.75}{728.5\times 2.2}\approx \frac{323.75}{1602.7}\approx 0.202$$

---

Final Answer:

The coefficient of kinetic friction is ${\mu }_k=0.202$ (to three significant figures).