One side of the roof of a building slopes up at 38.0°. A roofer kicks a round, f

Question

One side of the roof of a building slopes up at 38.0°. A roofer kicks a round, flat rock that has been thrown onto the roof by a neighborhood child. The rock slides straight up the incline with an initial speed of 15.0 m/s. The coefficient of kinetic friction between the rock and the roof is 0.390. The rock slides 10.0 m up the roof to its peak. It crosses the ridge and goes into free fall, following a parabolic trajectory above the far side of the roof, with negligible air resistance. Determine the maximum height the rock reaches above the point where it was kicked.

Explanation / Answer

Let the mass of the rock = m.

Force between rock and roof due to rock’s weight = F = mg.cos= mg.cos 38º

KE = 1/2 mv² = 1/2 * M * (15m/s)² = 112.5m^2/s^2 * M
work done by friction W = µmgcos*d = 0.390 * M * 9.8m/s² * cos38º * 10m = M*30.11m²/s²
increase in PE = mgh = M * 9.8m/s² * 10m * sin38º = M *60.33m²/s²
Then the rock has KE = M(112.5 - 30.11- 60.33)m²/s² = 22.06m²/s² * M = ½Mv²
so v =6.64 m/s
We have the launch angle and velocity:
max. height = (V·sin)² / (2g) = (6.64m/s * sin38)² / 19.6m/s² = 0.852m
plus the height it gained along the roof = 10m * sin38 = 6.156 m
yields a total rise h = 6.156+ 0.852m = 7.008m

maximum height the rock reaches = 7.008 m

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.