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Answer :
The sled moves approximately 5.17 meters up the incline.
Given:
Initial speed (vi) = 4.3 m/s
Mass of the monkey and sled (m) = 22.5 kg
Incline angle (θ) = 24 degrees
Coefficient of kinetic friction (μk) = 0.45
Acceleration due to gravity (g) = 9.8 m/s²
Determine the height gained (h) using the incline angle:
h = (vi^2 * sin^2(θ)) / (2 * g)
h = (4.3^2 * sin^2(24)) / (2 * 9.8)
h ≈ 1.018 m
Calculate the work done by gravity (Wg):
Wg = mgh
Wg = 22.5 * 9.8 * 1.018
Wg ≈ 225.747 J
Find the work done against friction (Wf):
Wf = -μk * mgh
Wf = -0.45 * 22.5 * 9.8 * 1.018
Wf ≈ -102.196 J
Apply the work-energy principle:
Wnet = (1/2) * m * vf^2 - (1/2) * m * vi^2
Since vf is 0 (sled comes to rest), the equation becomes:
Wg + Wf = -(1/2) * m * vi^2
Substitute values and solve for vi.
Now, use the formula for distance traveled up the incline (d):
d = (vf^2 - vi^2) / (2 * g) + μk
d = (0 - 4.3^2) / (2 * 9.8) + 0.45
d ≈ 5.17 m
Therefore, the sled moves approximately 5.17 meters up the incline.
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