Determine the acceleration with which a sled is rolling down a slide 7 m high and 17 m long. The coefficient of friction of the sleigh on the snow is 0.3.
h = 7 m.
L = 17 m.
μ = 0.3.
g = 10 m / s2.
m * a = Ft + Ftr + N, where Ft is the force of gravity that acts on the sled, Ftr is the friction force of the sled against the slide, N is the reaction force of the slide to the sled.
ОХ: m * a = m * g * sinα – Ftr.
OU: 0 = – m * g * cosα + N.
a = (m * g * sinα – Ftr) / m.
N = m * g * cosα.
Ftr = μ * N = μ * m * g * cosα.
a = (m * g * sinα – μ * m * g * cosα) / m = g * (sinα – μ * cosα).
sinα = h / L = 7 m / 17 m = 0.41.
cosα = √ (1 – sin2α).
cosα = √ (1 – (0.41) 2) = 0.91.
a = 10 m / s2 * (0.41 – 0.3 * 0.91) = 1.37 m / s2.
Answer: the sled will move with acceleration a = 1.37 m / s2.
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