# Answer to Question #1674 in Mechanics | Relativity for Brian

Question #1674

A child of mass 30 kg skates down from the top of a ramp having a constant slope of 15°. The child’s speed increases from 1.5 ms−1 to 3.0 ms−1 as she reaches the bottom of the ramp. A force of kinetic friction of magnitude 50 N opposes her motion. Determine the length of the ramp. Take g = 10 ms−2.

Expert's answer

The moving force is mg*sin 15 = 300 sin15 = 77.65 N, the resultative force that causes the motion downward is 77.65 - 50= 27.65 N. The

acceleration is F/m = 27.65/30 =0.92 m/s

According equations of the motion

h = h

v = v

t = (v-v0)/a = (3-1.5)/0.92 = 1.63 s.

Thus h = 1.5*1.63 + 0.92* 1.63

acceleration is F/m = 27.65/30 =0.92 m/s

^{2}.According equations of the motion

h = h

_{0}+ v_{0}t + at^{2}/2v = v

_{0}+ at, where v = 3m/s, v_{0}= 1.5 m/s, h_{0}= 0, a = 0.92 m/s^{2}.t = (v-v0)/a = (3-1.5)/0.92 = 1.63 s.

Thus h = 1.5*1.63 + 0.92* 1.63

^{2}/2 = 3.67 m.Need a fast expert's response?

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