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4 months ago

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In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it start

s 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.

2 answers:

Ostrovityanka [3.2K]4 months ago

8 0

Answer:

Explanation:

Applying trigonometry to calculate the slope's height:

sinθ = opposite / hypotenuse

sin 4° = h / 100

h = 100 × sin 4°

h = 6.98 m

Using conservation of energy principles:

Change in potential energy equals change in kinetic energy,

∆PE = ∆KE

mg(hf - hi) = ½ m (vf² - vi²)

The initial speed is:

vi = 9.2 m/s

Initial height is:

hi = 6.98 m

Final height when the sled reaches the ground:

hf = 0 m

We solve for final velocity vf:

mg(hf - hi) = ½ m (vf² - vi²)

Mass cancels out:

g(hf - hi) = ½ (vf² - vi²)

9.81 × (0 - 6.98) = ½ (vf² - 9.2²)

-68.43 = ½ (vf² - 84.64)

Multiplying both sides by 2:

-136.86 = vf² - 84.64

vf² = -136.86 + 84.64 = -52.22 (This is inconsistent; correcting the sign)

Actually, hf - hi = 0 - 6.98 = -6.98, but since the sled is descending, potential energy loss is positive, so flip the signs:

g(hi - hf) = ½ (vf² - vi²)

9.81 × 6.98 = ½ (vf² - 9.2²)

68.43 = ½ (vf² - 84.64)

136.86 = vf² - 84.64

vf² = 136.86 + 84.64 = 221.5

vf = √221.5 = 14.88 m/s

Maru [3.3K]4 months ago

3 0

Answer:

v = 14.87 m/s

Explanation:

Given:

Initial velocity, u = 9.2 m/s

Angle θ = 4°

Distance traveled, d = 100 m

Let v be the speed after traveling 100 m.

From the figure:

Height, h = d × sinθ

h = 100 × sin 4°

h = 6.97 m

Using physics equations,

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

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

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

Therefore, v = 14.87 m/s

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