Respuesta :
Answer:
She travels 6.94 m before stopping.
Explanation:
Given:
Initial velocity of the skier (u) = 5.47 m/s
Final velocity of the skier (v) = 0 m/s (Comes to stop)
Coefficient of kinetic friction (μ) 0.220
Let the mass of skier be 'm', acceleration of the skier be 'a' and the distance traveled be 'd' before coming to a stop.
Now, using the equation of motion as:
[tex]v^2=u^2+2ad[/tex]
Plugging in the given values and expressing 'a' in terms of 'd', we get:
[tex]0=(5.47)^2+2ad\\\\2ad=-29.92\\\\a=-\frac{29.92}{2d}=-\frac{14.96}{d}------(1)[/tex]
Now, the above acceleration is due to the frictional force. Now, frictional force is given as:
[tex]f=\mu N\\\\f=\mu mg\\\\f=0.220\times m\times 9.8=2.156m[/tex] -------------- (2)
Here, 'N' is the normal force which is equal to the weight of the skier.
From Newton's second law, frictional force is equal to the product of mass and acceleration. Here, friction acts in the opposite direction to motion.
So, [tex]f=-ma=m(\frac{14.96}{d})[/tex] ---------- (3)
Equating equations (2) and (3), we get:
[tex]2.156m=\frac{14.96m}{d}\\\\d=\frac{14.96}{2.156}=6.94\ m[/tex]
Therefore, she travels 6.94 m before stopping.
She travels 6.94 m on the patch before stopping.
Work energy theorem:
According to the work-energy theorem, the work done (W) by a force is equal to the change in kinetic energy (ΔKE) of the object:
W = ΔKE
The work done by frictional force (F) will be:
W = Fd
where d is the traveled distance
The friction force is given by
F = μmg
where μ = 0.220 is the coefficient of kinetic friction.
Thus, the work done by friction is:
W = μmgd
Now, the change in kinetic energy is :
[tex]\Delta KE = \frac{1}{2}m(\Delta v)^2[/tex]
Applying work-energy theorem:
W = ΔKE
[tex]\mu mgd= \frac{1}{2}m(\Delta v)^2\\\\d=\frac{(\Delta v)^2}{2\mu g}\\\\d=\frac{(5.47)^2}{2\times0.220\times9.8}\\\\d= 6.94\;m[/tex]
She travels 6.94 m before stopping.
Learn more about work-energy theorem:
brainly.com/question/13603991?referrer=searchResults
