A skier slides horizontally along the snow for a distance of 21 m before coming to rest. The coefficient of kinetic friction between the stier and the snow is `mu_(k)=0.050`. Initially, how fast was the skier going ?

Options

A. 6.4m/s
B. 4.5m/s
C. 2.7m/s
D. 5.4m/s

Answer

To solve the problem, we will use the concepts of work, energy, and friction. Here’s a step-by-step solution:

### Step 1: Understand the forces acting on the skier
The skier is sliding on snow, and the only horizontal force acting on the skier is the frictional force, which opposes the motion. The coefficient of kinetic friction (\( \mu_k \)) is given as 0.050.

### Step 2: Write down the expression for the frictional force
The frictional force (\( F_f \)) can be calculated using the formula:
\[ F_f = \mu_k \cdot m \cdot g \]
where:
- \( \mu_k = 0.050 \) (coefficient of kinetic friction)
- \( m \) is the mass of the skier (which will cancel out later)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

### Step 3: Calculate the work done by friction
The work done by friction (\( W_f \)) as the skier slides a distance \( s \) is given by:
\[ W_f = -F_f \cdot s \]
Substituting the expression for frictional force:
\[ W_f = -(\mu_k \cdot m \cdot g) \cdot s \]

### Step 4: Relate work done to change in kinetic energy
According to the work-energy principle:
\[ W_f = \Delta KE = KE_f - KE_i \]
Since the skier comes to rest, the final kinetic energy (\( KE_f \)) is 0. The initial kinetic energy (\( KE_i \)) is given by:
\[ KE_i = \frac{1}{2} m u^2 \]
Thus, we have:
\[ -\mu_k \cdot m \cdot g \cdot s = 0 - \frac{1}{2} m u^2 \]

### Step 5: Simplify the equation
Cancelling \( m \) from both sides (since \( m \) is not zero):
\[ -\mu_k \cdot g \cdot s = -\frac{1}{2} u^2 \]
Rearranging gives:
\[ \frac{1}{2} u^2 = \mu_k \cdot g \cdot s \]

### Step 6: Solve for initial velocity \( u \)
Multiplying both sides by 2:
\[ u^2 = 2 \cdot \mu_k \cdot g \cdot s \]
Substituting the known values:
- \( \mu_k = 0.050 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( s = 21 \, \text{m} \)

Calculating:
\[ u^2 = 2 \cdot 0.050 \cdot 10 \cdot 21 \]
\[ u^2 = 2 \cdot 0.050 \cdot 210 \]
\[ u^2 = 21 \]

Taking the square root:
\[ u = \sqrt{21} \approx 4.58 \, \text{m/s} \]

### Final Answer
The initial speed of the skier was approximately \( 4.58 \, \text{m/s} \).

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