HW 6 Solns F19.pdf - Physics 1112 Homework#6 Solutions#6.55[Ski Tow Power Neglect friction Find the power needed by calculating the rate of increase of

HW 6 Solns F19.pdf - Physics 1112 Homework#6...

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Physics 1112 Homework #6 Solutions #6.55 [Ski Tow Power] Neglect friction. Find the power needed by calculating the rate of increase of total mechanical energy of the skiers. Since they move with constant speed, their KE doesn't change. Only their gravitational PE changes as the height y of each skier increases at: . Let M = the total mass of all the skiers = (50)(70.0 kg) = 3.50 x 10 3 kg. Their speed v = (12.0 km/h) = (12.0 x 10 3 m/h)(1 h/3600 s) = 3.33 m/s Power: = Mg v y = Mg v sin q = (3.50 x 10 3 kg)(9.80 N/kg)(3.33 m/s)sin15.0° = 29.6 kW = (29.6 kW)(1 hp/0.746 kW) = 39.6 hp OR Another way to find the power is to use , where F || is the component parallel to the velocity of the force pulling the skiers up the hill or the force due to the ski tow. Newton's 2nd Law: +x: F || - Mg sin q = ma x = 0 [Constant v] Þ F || = Mg sin q = (3.50 x 10 3 kg)(9.80 N/kg)sin15° = 8.88 x 10 3 N Power: P = F||v = (Mg sin q ) v = Mg v sin q , as before. #7.76 [PE Graph] (a) . The slope of the PE graph is < 0 at A, so F x (A) > 0. The force is in the +x direction ( ® ), pushing the particle towards lower PE and larger x. (b) At point B the slope of the PE graph is > 0, so F x (B) < 0. The force on the particle is pointing in the -x direction ( ¬ ), pushing the particle to smaller x and lower PE. (c) KE is maximum when PE is minimum: around 0.75 m. (d) The curve at point C looks pretty close to horizontal (slope = 0), so the force there is zero. (e) The object had zero KE at point A, so it can only reach a point where the PE equals the PE at A (like a pendulum swinging back and forth between positions of equal height). On the graph, this looks like about 2.2 m, labeled point D and indicated by the dashed line at the PE height of A. (f) The points of local minimum PE: the one found in part (c) at x = 0.75 m and the other one around 1.9 m. (g) The only PE local maximum: point C. (h) The particle is moving slowest and not changing direction at point C. Although it is momentarily at rest at points A and D, it is changing direction there, so these are "turning points" of its motion. dy dt = v y = v sin θ P = dU g dt = d dt ( Mgy ) = Mg dy dt P = ! F ! v = F || v F x = dU dx dU dx v L θ M +y +x θ F || N Mg v D
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#13.51 [Escape Speed] Diameter = 300 km = 3.00 x 10 5 m, Radius = 1.50 x 10 5 m Density r = M/V , with for a sphere. Escape speed: = = 177 m/s This is much less than the escape speed for the Earth: 11.2 km/s = 11200 m/s. #13.69 [Orbit Energies] Radius of Mars: R = 3.39 x 10 6 m Mass of Mars: M = 6.42 x 10 23 kg Orbit #1 radius: r 1 = 2000 x 10 3 m + 3.39 x 10 6 m = 5.39 x 10 6 m Orbit #2 radius: r 2 = 4000 x 10 3 m + 3.39 x 10 6 m = 7.39 x 10 6 m To find the circular orbit speed, use Newton's 2nd Law: Þ Þ Orbit #1: Spacecraft's KE: Gravitational PE: Total mechanical energy in orbit #1: E 1 = K 1 + U g1 = Similarly, total mechanical energy in orbit #2: E 2 = K 2 + U g2 = Work to move from orbit #1 into orbit #2: W = E 2 - E 1 = = = 5.38 x 10 9 J V = 4 3 π R 3 v = 2 GM R 2 G ρ (4 π R 3 / 3) R = R 8 π G ρ 3 = (1.50 × 10 5 m ) 8 π (6.67 × 10 11 N - m 2 / kg 2 )(2500 kg / m 3 ) 3 F = m a GM E m r 2 = m v 2 r v = GM E r K 1 = 1 2 mv 1 2 = 1 2 m GM r 1 = GMm 2 r 1 U g 1 = GMm r 1 GMm 2 r 1 GMm r 1 = GMm 2 r 1 GMm 2 r 2 GMm r 2 = GMm 2 r 2 GMm 2 r 2 GMm 2 r 1 = GMm 2 1 r 1 1 r 2 (6.67 × 10 11 N - m 2 / kg 2 )(6.42
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