Questions: Problem 1: A 100 kg man is skiing across level ground at a speed of 8.0 m / s when he comes to the small slope 1.8 m higher than ground level shown in the following figure. (a) If the skier coasts up the hill, what is his speed when he reaches the top plateau? Assume friction between the snow and skis is negligible. (b) What is his speed when he reaches the upper level if an 80 N constant frictional force acts on the skis?

Problem 1: A 100 kg man is skiing across level ground at a speed of 8.0 m / s when he comes to the small slope 1.8 m higher than ground level shown in the following figure.
(a) If the skier coasts up the hill, what is his speed when he reaches the top plateau? Assume friction between the snow and skis is negligible.
(b) What is his speed when he reaches the upper level if an 80 N constant frictional force acts on the skis?

Transcript text: Problem 1: A 100 kg man is skiing across level ground at a speed of $8.0 \mathrm{~m} / \mathrm{s}$ when he comes to the small slope 1.8 m higher than ground level shown in the following figure. (a) If the skier coasts up the hill, what is his speed when he reaches the top plateau? Assume friction between the snow and skis is negligible. (b) What is his speed when he reaches the upper level if an 80 N constant frictional force acts on the skis?

Solution

Solution Steps

Step 1: Analyze the Problem and Apply Conservation of Energy

For part (a), we assume there is no friction. The skier's initial kinetic energy is converted into potential energy as he ascends the slope. The principle of conservation of energy states:

\[ \text{Initial Kinetic Energy} = \text{Final Kinetic Energy} + \text{Potential Energy at the top} \]

The initial kinetic energy ($KE_i$) is given by:

\[ KE_i = \frac{1}{2} m v_i^2 \]

where $m = 100 \, \text{kg}$ and $v_i = 8.0 \, \text{m/s}$.

The potential energy ($PE$) at the top is:

\[ PE = mgh \]

where $g = 9.81 \, \text{m/s}^2$ and $h = 1.8 \, \text{m}$.

The final kinetic energy ($KE_f$) is:

\[ KE_f = \frac{1}{2} m v_f^2 \]

where $v_f$ is the final speed at the top.

Step 2: Solve for Final Speed without Friction

Using conservation of energy:

\[ \frac{1}{2} m v_i^2 = \frac{1}{2} m v_f^2 + mgh \]

Cancel $m$ from each term:

\[ \frac{1}{2} v_i^2 = \frac{1}{2} v_f^2 + gh \]

Rearrange to solve for $v_f$:

\[ v_f^2 = v_i^2 - 2gh \]

Substitute the known values:

\[ v_f^2 = (8.0)^2 - 2 \times 9.81 \times 1.8 \]

\[ v_f^2 = 64 - 35.316 \]

\[ v_f^2 = 28.684 \]

\[ v_f = \sqrt{28.684} \approx 5.355 \, \text{m/s} \]

Step 3: Analyze the Problem with Friction

For part (b), we include the work done by friction. The work done by friction ($W_f$) is:

\[ W_f = f \cdot d \]

where $f = 80 \, \text{N}$ and $d$ is the distance along the slope. However, we need to find $d$ using the height and the angle of the slope, which is not given. Assuming the slope is small, we can approximate $d$ using the height:

\[ d \approx \frac{h}{\sin(\theta)} \approx h \]

Thus, the work done by friction is approximately:

\[ W_f = 80 \times 1.8 = 144 \, \text{J} \]

Step 4: Solve for Final Speed with Friction

The energy equation with friction becomes:

\[ \frac{1}{2} m v_i^2 = \frac{1}{2} m v_f^2 + mgh + W_f \]

Cancel $m$ from each term:

\[ \frac{1}{2} v_i^2 = \frac{1}{2} v_f^2 + gh + \frac{W_f}{m} \]

Rearrange to solve for $v_f$:

\[ v_f^2 = v_i^2 - 2gh - \frac{2W_f}{m} \]

Substitute the known values:

\[ v_f^2 = (8.0)^2 - 2 \times 9.81 \times 1.8 - \frac{2 \times 144}{100} \]