Answer:
Sam's top speed = 25 m/s
Distance Sam travels before coastring to a stop = 360 m
Explanation:
The thrust from the skis = 250 N
Frictional force on Sam = coefficient of friction × Sam's weight
= (0.1)(71 kg)(9.8 m/s²) = 69.58 N
The effective force moving Sam = 250 N - 69.58 N = 180.42 N
The acceleration of Sam due to this force = 180.42 N ÷ 71 kg = 2.54 m/s²
Sam starts from rest. Initial velocity, <em>u</em> = 0 m/s. The skis run out after time, <em>t</em> = 10 s.
Using the equation of motion,
<em>v</em> = <em>u</em> + <em>at</em>
<em>v</em> = 0 + (2.54 m/s²)(10 s) = 25.4 m/s ≈ 25 m/s
This is Sam's top speed.
<em>v</em>² = <em>u</em>² + 2<em>as</em>
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The distance travelled while the skis were working is determined using the equation of motion
For the stopping motion, his initial velocity is 25.4 m/s and final velocity = 0 m/s. His acceleration is determined by the frictional force since there is no thrust from the skis.
<em>a</em> = -69.58 N ÷ 71 kg = -0.98 m/s²
(It is negative because it acts in opposite direction to the velocity)
Using the equation of motion
<em>v</em>² = <em>u</em>² + 2<em>as</em>
![s = \dfrac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=s%20%3D%20%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
![s = \dfrac{(0\text{ m/s})^2-(25.4\text{ m/s})^2}{2\times(-0.98\text{ m/s}^2)} = 329.16\text{ m}](https://tex.z-dn.net/?f=s%20%3D%20%5Cdfrac%7B%280%5Ctext%7B%20m%2Fs%7D%29%5E2-%2825.4%5Ctext%7B%20m%2Fs%7D%29%5E2%7D%7B2%5Ctimes%28-0.98%5Ctext%7B%20m%2Fs%7D%5E2%29%7D%20%3D%20329.16%5Ctext%7B%20m%7D)
Total distance travelled = 32.92 m + 329.16 m = 362.08 m ≈ 360 m