Question

In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it starts to careen down the steeply sloped track. In one event, the sled reaches a top speed of 9.2 m/s before starting down the initial part of the track, which is sloped downward at an angle of 4.0 degrees.What is the sled's speed after it has traveled the first 100 m? Express your answer with the appropriate units.

Answer 1

Answer:

Explanation:

Using trigonometric to find the height of the slope

Sinθ = opp / hyp

Sin 4 = h / 100

h = 100×Sin4

h = 6.98 m

So, using conservation of energy

Change Potential energy = change in kinetic energy

∆ P.E = ∆K.E

mg(hf-hi) = ½m(vf²-vi²)

The, initial velocity is given as

Vi = 9.2m/s

The initial height is

hi = h = 6.98m

The final height is when the body hits the ground and at that point, the height is 0

hf = 0m

Then we need to find the final velocity cf

mg(hf-hi) = ½m(vf²-vi²)

m cancels out

g(hf-hi) = ½(vf²-vi²)

9.81 ( 6.98 - 0) = ½(vf²-9.2²)

68.43 = ½(vf²-9.2²)

68.43 × 2 = vf²-9.2²

vf² = 68.43 × 2 + 9.2²

vf² = 221.50

vf = √221.5

vf = 14.88m/s

Answer 2

Answer:

v = 14.87 m/s

Explanation:

Given that

Initial speed ,u= 9.2 m/s

θ = 4°

Distance d= 100 m

Lets speed after traveling 100 m is v.

From the diagram

h = d sinθ

h = 100 sin 4°

h = 6.97 m

We know that

$v^2=u^2+2ah$

$a=g=9.81\ m/s^2$

$v^2=9.2^2+2\times 9.81\times 6.97$

v = 14.87 m/s