Respuesta :
Explanation:
It is given that,
Mass of the ice skater, m = 50 kg
Initial speed of the car, u = 4 m/s
The force is exerted due north east direction, F = 4 N
Angle, [tex]\theta=45^{\circ}[/tex]
(a) Let v is the speed of the skater after gliding 100 m in this wind. Using the work energy theorem to find it :
[tex]\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=W[/tex]
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}mu^2+Fd\ cos(45)[/tex]
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}\times 50\times (4)^2+4\times 100\times cos(45)[/tex]
[tex]v=\sqrt{\dfrac{2\times 682.84}{50}}[/tex]
v = 5.22 m/s
(b) Let [tex]\mu_s[/tex] is the coefficient of static friction that allows her to continue moving straight north. It is given by :
[tex]f=\mu N[/tex][tex]\mu=\dfrac{f}{N}[/tex]
[tex]\mu=\dfrac{4\ cos(45)}{50\times 9.8}[/tex]
[tex]\mu=0.0057[/tex]
Hence, this is the required solution.
The minimum value of [tex]\mu_k[/tex] that allows her to continue moving straight north is 0.0057.
What is friction force?
The friction force is the force that is applied on an object which is opposite in direction of the movement of the object.
As we know that the work is the change in the kinetic energy of the system, therefore, the skater’s speed after gliding 100 m in this wind can be written as,
[tex]W = \dfrac{1}{2} m(v^2-u^2)[/tex]
As the wind is coming from the northeast, therefore, the angle made by the force is 45°, therefore,
[tex]F\cdot cos(45^o)\cdot d= \dfrac{1}{2} m(v^2-u^2)\\\\4\times \dfrac{1}{\sqrt{2}}\times 100= \dfrac{1}{2}\times 50 \times (v^2-4^2)\\\\v = 5.22\rm\ m/s[/tex]
Hence, the skater’s speed after gliding 100 m in this wind is 5.22 m/s.
B.) The minimum value of [tex]\mu_k[/tex] that allows her to continue moving straight north
In order to find the minimum value of [tex]\mu_k[/tex] that allows her to continue moving straight north,
Friction force, F = μN
Substitute the values,
[tex]F = \mu N\\\\4\ cos(45^o)=\mu \times mg\\\\4\ cos(45^o) = \mu \times 50 \times 9.81\\\\\mu_k= 0.0057[/tex]
Hence, the minimum value of [tex]\mu_k[/tex] that allows her to continue moving straight north is 0.0057.
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