Respuesta :
Answer:
The speed of her at the bottom is 24.035 m/s.
Explanation:
Given that,
Speed of cyclist = 12 m/s
Height = 30 m
Distance d = 450 m
Mass of cyclist and bicycle =70 kg
Drag force = 12 N
We need to calculate the speed at the bottom
Using conservation of energy
K.E+P.E=drag force+K.E+P.E
Potential energy is zero at the bottom.
K.E+P.E=drag force+K.E
[tex]\dfrac{1}{2}mv^2+mgh=Fx+\dfrac{1}{2}mv^{2}[/tex]
[tex]\dfrac{1}{2}\times70\times12^2+70\times9.8\times30=12\times450+\dfrac{1}{2}\times70\timesv^2[/tex]
[tex]25620=5400+35v^2[/tex]
[tex]35v^2=25620-5400[/tex]
[tex]35v^2=20220[/tex]
[tex]v^2=\dfrac{20220}{35}[/tex]
[tex]v=\sqrt{\dfrac{20220}{35}}[/tex]
[tex]v=24.035\ m/s[/tex]
Hence, The speed of her at the bottom is 24.035 m/s.
Answer:
Speed at the bottom of the slope is 24.03 m /sec
Explanation:
We have given speed of the cyclist v = 12 m/sec
She starts down a 450 m long that is 30 m high
Si distance = 450 m and height h = 30 m
Combined mass of cyclist and bicycle m = 70 kg
Drag force = 12 N
We have to find the speed at the bottom
According to conservation theory
KE +PE = work done by drag force + KE + PE
As we know that at the bottom there will be no potential energy
So [tex]\frac{1}{2}\times 70\times 12^2+70\times 9.8\times 30=12\times 450+0+\frac{1}{2}\times 70\times v^2[/tex]
[tex]\frac{1}{2}\times 70\times v^2=20220[/tex]
[tex]v=24.03m/sec[/tex]
