Problem_Set_7

Anthony Putrino Problem Set #7 - Problems 10, 20, 28, 40 73 10. A 57-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37° above the horizontal. If the tension in the rope is 142 N, how much work is done on the crate to move it 6.1 m? Given: T=142N, θ=37°, d = 6.1mandm = 57kg Work done is W= ´ T . ´ d =T.d.cosθ =142 × 6.1×cos37° =) W= 691.78J * Work done on crate is work done by tension force. Final answer: Required work done is W = 691.78 Joules 20. A 9.50-g bullet has a speed of 1.30 km/s, (a) What is the kinetic energy in joules? (b) What is the bullet’s kinetic energy if its speed is halved? (c) If its speed is doubled? Given: mass = 9.50 g = 0.0095 kg . speed = 1.30 km/s = (1.30 * 1000) m/s = 1300 m/s . (a) kinetic energy K1 = 1 2 mv 2 = 1 2 × 0.0095×1,300 2 K1 = 8027.5 J . (b) now kinetic energy of bullet if speed is halved . k2 = 1 2 m ( v 2 ) 2 = 1 2 × 0.0095×( 1,300 2 ) 2 K2 = 2006.875 J . (c) now kinetic energy of bullet if speed is doubled . K3 = 1 2 m(2v) 2 = 1 2 × 0.0095×(2×1,300) 2 K3 = 32110 J . Final answer: (a) K1 = 8027.5 J . (b) K2 = 2006.875 J . (c) K3 = 32110 J .

28. A 1100 kg. car coasts on a horizontal road with a speed of 19 m/s. After crossing an unpaved, sandy stretch of road 32 m long, its speed decreases to 12 m/s. (a) Was the net work done on the car positive, negative or zero? Explain. (b) Find the magnitude of the average net force on the car in the sandy section. Solution:-Using equation of motion; v 2 = u 2 + 2as 12 2 =19 2 +2 × a× 32 a = −3.39 m s 2 negative sign means deceleration. Now, using Newton's second law; F net = ma=1,100 × (−3.39) =−3,729N magnitude=3,729N *Here, first equation of motion is used and then Newton's second law is used i.e. v 2 = u 2 + 2as and F net = ma Answer: 3,729N 40. A new record for running the stairs of the Empire State Building was set on February 4, 2003. Thee 86 flights, with a total of 1576 steps , was run in 9 minutes and 33 seconds. If the height gain of each step was 0.20 m, and the mass of the runner was 74 kg., what was his average power output during the climb? Give your answer in both watts and horsepower. Solve for power and convert to watts and/or horsepower. P = W/t = (74kg)(9.81m/s 2 )(1576 steps*.20m)/(9min*60s/min+33sec) = 216447.84/573s/min = 377.744 W Covert watts to horsepower.