Third physics Exam

(a)Find the useful power output (in W) of an elevator motor that lifts a 2300 kg load a height of 25.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg—so that only 2300 kg is raised in height, but the full 10,000 kg is accelerated. (b) What does it cost (in cents), if electricity is $0.0900 per kW · h?

(a) W = KE + PE W = 1/2 mv^2 + mgh P = w/t P = 1 / t(mv^2 / 2 + mgh) P = 1 / 12((10,000 kg)(4)^2 / 2 + (2,300 kg)(9.8 m/s^2)(25)) (NOTE FOR MYSELF: Enter this last part EXACTLY as it is shown and input it EXACTLY in the calculator.)

A 50.0 kg skier with an initial speed of 14.0 m/s coasts up a 2.50 m high rise as shown in the following figure. Find her final speed at the top (in m/s), given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)

-mgh - (umg costheta) L = 1/2 mv^2 - 1/2 mv^2 -gh - (ug costheta) L = 1/2 v^2lowerf - 1/2 v^2loweri vf = square root of v^2loweri - 2g(h + uL costheta) vf = square root of v^2loweri - 2g(h + u(h / sintheta) costheta) vf = square root of v^2loweri - 2gh (1 + u / tan theta) vf = square root of (14)^2 - 2(9.8 m/s^2)(2.5 m) (1 + 0.08 / tan 35) vf = 11.9 m/s (NOTE FOR MYSELF: The 35 came from the drawing in the homework.)

A 7.00 ✕ 10^5 kg subway train is brought to a stop from a speed of 0.500 m/s in 0.800 m by a large spring bumper at the end of its track. What is the force constant k of the spring (in N/m)?

1/2 kx^2 = -1/2 mvf^2 + 1/2 mv0^2 1/2 kx^2 = -0 + 1/2 mv0^2 k = mv0^2 / x^2 K = (7.00 * 10^5) (0.5 m/s)^2 / 0.8^2 273437.5

In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy even on small hills.) To demonstrate this, find the final speed in m/s and the time taken in seconds for a skier who skies 69.0 m along a 27° slope neglecting friction for the following two cases. (Enter the final speeds to at least one decimal place.) (a) starting from rest (b) starting with an initial speed of 3.50 m/s

A. (mg sin theta) L = 1/2 mv^2 - 1/2 mv^2 The initial velocity is zero. (mg sin theta) L = 1/2 mv^2 - 0 vf = square root of 2(9.8 m/s^2)sin27(69 m) vf = 24.7 m/s vf - vi = at t = vg - vi/a t = 24.77 m/s / (9.8 m/s^2) sin27 t = 5.56 s B. B. (mg sin theta) L = 1/2 mv^2 - 1/2 mv^2 The initial velocity is zero. 2(gsin theta) L = v^2 - v^2 vf = square root of v^2 + 2(gsin theta) L vf = square root of (3.5 m/s)^2 + 2(9.8 m/s^2) sin27 (69 m) vf = 25 m/s t = vf - vi / a t = 25 m/s - 3.5 m/s / (9.8 m/s^2) sin27 t = 4.83 s

(a) Calculate the force (in N) needed to bring a 800 kg car to rest from a speed of 95.0 km/h in a distance of 105 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).

A. 95 km/h * 1,000 / 3,600 = 26.4 m/s a = V^2 / 2s a = (26.4 m/s)^2 / 2 * 105 = 3.31 m/s^2 F = ma 800 kg * 3.31 m/s^2 2655 N B. a = v^2 / 2s a = (95 * 5 / 18)^2 / 2(2) a = 174.09 m/s^2 F = ma F = (800 kg)(174.09) F = 139274.7 force in (b) / force in (a) = 139274.7 N / 2655 N 52.45

(a) How much gravitational potential energy (in J) (relative to the ground on which it is built) is stored in an Egyptian pyramid, given its mass is about 3 ✕ 10^9 kg and its center of mass is 17.0 m above the surrounding ground? (b) What is the ratio of this energy to the daily food intake of a person (1.2 ✕ 10^7 J)?

A. MGH = (3*10^9)(9.8)(17) 4.998E11 B. 4.998E11 / 1.2*10^7 41650

(a) What is the mass of a large ship that has a momentum of 2.60 ✕ 10^9 kg·m/s, when the ship is moving at a speed of 42.0 km/h?

A. Mass = Pship / v 2.60 * 10^9 kg * m/s / 42 km/hr 2.60 * 10^9 kg * m/s / 42 (1,000 / 3600) m/s 2.60 * 10^9 kg * m/s / 11.6 m/s 224137931 kg

(a) What is the momentum in kg · m/s of a garbage truck that is 1.30 ✕ 10^4 kg and is moving at 35.0 m/s? (b) At what speed in m/s would an 8.00 kg trash can have the same momentum as the truck?

A. Momentum = mass * velocity 1.30 * 10^4 * 35 455,000 kg * m/s B. Velocity = power / mass 455,000 kg * m/s / 8 kg 56875 m/s

(a) Calculate the momentum of a 2100 kg elephant charging a hunter at a speed of 7.50 m/s. (b) Compare the elephant's momentum with that of a 0.0400 kg bullet fired at a speed of 600 m/s. (c) What is the momentum in kilograms times meters per second of the 90.0 kg hunter running at 7.00 m/s after missing the elephant?

A. Momentum = mass * velocity 2,100 kg * 7.5 m/s^2 = 15750 kg * m/s B. Momentum = mass * velocity Momentum of bullet = 0.04 kg * 600 m/s = 24 kg * m/s (1570 kg * m/s) / (24 kg * m/s) = 656.25 C. Momentum = mass * velocity Momentum of hunter = 90 kg * 7 m/s = 630 kg * m/s

A 0.850 kg hammer is moving horizontally at 5.50 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board. (a) Calculate the duration of the impact in seconds. (b) What was the average force in newtons exerted downward on the nail?

A. v^2 = u^2 + 2as 2as = v^2 - u^2 2a(0.01) = 0 - 5.5^2 0.02a = -30.25 a = -1512.5 u + at = v u + at = 0 5.5 - 1512.5t = 0 -1512.5t = -5.5 1512.5t = 5.5 t = 0.003636 s B. F = ma F = 0.850 kg * 1512.5 F = 1285.625 N

A runaway train car that has a mass of 14,000 kg travels at a speed of 4.5 m/s down a track. Compute the time (in s) required for a force of 1800 N to bring the car to rest.

Fnet = m triangle v / triangle t triangle t = m triangle v / Fnet triangle v = v2 - v1 v2 is the force that has brought the car to rest, so the final speed is 0 m/s triangle v = 0 - 4.5 m/s triangle v = -4.5 m/s triangle v = l -4.5 m/s l triangle v = 4.5 m/s triangle t = (14,000 kg) (4.5 m/s) / 1,800 N triangle t = 35 s

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 140,000 kg and a velocity of 0.300 m/s, and the second having a mass of 125,000 kg and a velocity of −0.120 m/s. (The minus indicates direction of motion.) What is their final velocity (in m/s)?

Pi = Pf m1v1 + m2v2 = (m1 + m2)vf vf = m1v1 + m2v2 / (m1 + m2) vf = (140,000 kg)(0.300 m/s) + (125,000 kg)(-0.130 m/s) / (125,000 kg + 140,000 kg) vf = 0.09716 m/s

A car moving at 20 m/s crashes into a tree and stops in 0.35 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 66 kg. (Enter the magnitude.)

V = u + at The final velocity is 0, so V = 0. 0 = 20 m/s + (ax0.35) 0.35a = -20 m/s a = -57.1 m/s^2 F = m * I a I F = 66 kg * 57.1 F = 3768.6

How much work does a supermarket checkout attendant do on a can of soup he pushes 0.850 m horizontally with a force of 5.50 N? Express your answer in joules and kilocalories.

W = Fd cos theta (5.5 N)(0.85 m) (cos 0) 4.765 J (4.765 J) (1 kcal / 4,186 J) .0011383182 kilocalories

A 700 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s?

W = fd W = 1,200 N *400 = 480,000 J Ek = 1/2 mv^2 Ek = 1/2 (700 kg) (100)^2 Ek = 3,500,000 J Total energy input = 480,000 J + 3,500,000 J = 3,980,000 J P = w/t P = 3,980,000 J / 7.3 s = 545205.4795 W Hp = 746 W 545205.4795 W / 746 = 730.8 hp

A 1.40 kg falcon catches a 0.140 kg dove from behind in midair. What is their velocity (in m/s) after impact if the falcon's velocity is initially 26.0 m/s and the dove's velocity is 2.00 m/s in the same direction? (Enter the magnitude.)

m1v1 + m2v2 = (m1 + m2)v v = m1v2 + m2v2 / (m1 + m2) v = (1.40 kg)(26 m/s) + (0.140 kg)(2 m/s) / (1.40 kg + 0.140 kg) v = 23.8 m/s