Answer:
Final speed, v = 12.57 m/s
Explanation:
Given that,
Slope with respect to horizontal, [tex]\theta=10.5^{\circ}[/tex]
Distance travelled, d = 19 m
Initial speed of the cyclist, u = 9.5 m/s
We need to find the final speed of the cyclist. Let h is the height of the sloping surface such that,
[tex]h=d\times sin\theta[/tex]
[tex]h=19\times sin(10.5)[/tex]
h = 3.46 m
Let v is the final speed of the cyclist. It can be calculated using work energy theorem as :
[tex]\dfrac{1}{2}m(v^2-u^2)=mgh[/tex]
[tex]\dfrac{1}{2}(v^2-u^2)=gh[/tex]
[tex]\dfrac{1}{2}\times (v^2-(9.5)^2)=9.8\times 3.46[/tex]
v = 12.57 m/s
So, the final speed of the 12.57 m/s. Hence, this is the required solution.
