Respuesta :
Explanation:
The given data is as follows.
Mass, m = 62 kg, Initial speed, [tex]v_{1}[/tex] = 6.90 m/s
Length of rough patch, L = 4.50 m, coefficient of friction, [tex]\mu_{k}[/tex] = 0.3
Height of inclined plane, h = 2.50 m
According to energy conservation equation,
(Final kinetic energy) + (Final potential energy) = Initial kinetic energy + Initial potential energy - work done by the friction
[tex]K.E_{2} + U_{2} = K.E_{1} + U_{1} - W_{f}[/tex]
[tex]\frac{1}{2}mv^{2}_{2} + U_{2} = \frac{1}{2}mv^{2}_{1} + mgh - \mu_{k}mgL[/tex]
Since, final potential energy is equal to zero. Therefore, the equation will be as follows.
[tex]\frac{1}{2}mv^{2}_{2} = \frac{1}{2}mv^{2}_{1} + mgh - \mu_{k}mgL[/tex]
Cancelling the common terms in the above equation, we get
[tex]\frac{1}{2}v^{2}_{2} = \frac{1}{2}v^{2}_{1} + gh - \mu_{k}gL[/tex]
= [tex]\frac{1}{2}(6.90)^{2}_{1} + 9.8 m/s^{2} \times 2.50 m - (0.3 \times 9.8 \times 4.50 m)[/tex]
= 36.055 - 13.23
= 22.825
[tex]v_{2} = \sqrt{2 \times 22.825}[/tex]
= 6.75 m/s
Thus, we can conclude that the skier is moving at a speed of 6.75 m/s when she gets to the bottom of the hill.
Answer:
Explanation:
mass, m = 62 kg
initial velocity, u = 6.9 m/s
length, l = 4.5 m
height, h = 2.5 m
coefficient of friction, μ = 0.3
Final kinetic energy + final potential energy = initial kinetic energy + initial potential energy + wok done by friction
Let the final velocity is v.
0.5 mv² + 0 = 0.5 mu² + μmgl + mgh
0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.5
0.5 v² = 23.805 + 13.23 + 24.5
v² = 123.07
v = 11.1 m/s
