Question

A boy of mass 30.0 kg is sledding down a 70.0-m slope starting from rest. The slope is angled at 15.0° below the horizontal. After going 28.0 m along the slope, he passes his friend, who hops onto the sled. The friend has a mass of 50.0 kg, and the coefficient of kinetic friction between the sled and the snow is 0.120. Ignoring the mass of the sled, find their speed at the bottom.
In m/s

Answer 1

The speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

Their speed at the bottom

Apply the principle of conservation of energy,

E(up) - E(friction) = E(bottom)

mg sin(15) + ¹/₂(M + m)u² - μ(M + m)cos 15 = ¹/₂(M + m)v²

$v = \sqrt{2[\frac{mgd \ sin15 \ + \frac{1}{2}(M + m)u^2 \ -\mu (M + m)g cos\ 15 }{M + m}] }$

where;

• u is the speed of the after 28 m

u = √2gh

u = √(2gL sin15)

u = √(2 x 9.8 x 28 x sin 15)

u = 11.92 m/s

$v = \sqrt{2[\frac{(30)(9.8)(70) \ sin15 \ + \frac{1}{2}(30 + 50)(11.92)^2 \ - 0.12 (30 + 50)9.8 cos\ 15 }{30 + 50}] }\\\\v = 16.52 \ m/s$

Thus, the speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

Learn more about speed here: brainly.com/question/6504879

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