The Expert TA l Human-like Grading, Automated! Home Student: kfiankoo uncc.edu C

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The Expert TA l Human-like Grading, Automated! Home Student: kfiankoo uncc.edu Chapter 4-B test Begin Date: 6/3/2017 12 0000 PM- Due Date: 6/5/2017 11:59:00 PM End Date: 6/28/2017 12:59:00 AM is (0%) Problem 7: Consider a skier heading down a 9.5e slope. Assume the coefficient of friction for waxed wood on wet snow 0.10 and use a coordinate system in which down the slope is positive. Randomized Variables 50% Part (a) Calculate the acceleration of the skier in m/s constant velocity in degrees. 50% Part (b) Find the angle of the slope down which this skier could coast at a Grade Summary 15 Deductions Potential 85 Submissions 7 8 9 Attempts remaining coso tan 0 sin0 (50 per attempt cotan0 asino acos0 detailed view atano acotano sinh0 1 2 3 cotanho Degrees Radians I give up! Hint Feedback: 5 deduction per feedback. Hints: 5% deduction per hint. Hints remaining Au content 2017 Expert TAuc

Explanation / Answer

Normal force of the slope on the skier is
Fn = mgcos

Friction force is
Ff = Fn
Ff = mgcos

Which is the force restricting acceleration down hill

The force accelerating the skier down hill is

Fa = mgsin

The net force is

F = Fa - Ff
F = mgsin - mgcos
F = mg(sin - cos)

From Newton

F = ma
we can then see that the net acceleration is
a = g(sin - cos)

a = 9.81(sin9.5 - 0.10cos9.5)
a = 0.65 m/s²

at constant velocity the acceleration equals zero

a = g(sin - cos)
0 = g(sin - cos)
as g is assumed to be not zero, then
0 = sin - cos
sin = cos
sin/cos = cos/cos
tan =
tan = 0.10
= 5.71°

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