University:
Jackson College
Course:
PHY 231 | College Physics I
Academic year:

2021
Views:

314

Pages:

4
Author:

Reginald Stewart

**Work, Energy & Power Worksheet** **1. If the cart goes from the bottom to the top in 60 seconds, how much power was developed?** 15m /| / | / | / | /____| 6.0 m mass 40 kg, no friction P = W/t P = (40 kg)(9.8 m/s²)(6.0 m) / 60 s P = 39 W **2. A 12 kW motor lifts a 1500 kg car 5.0 m in 15 s. Calculate the efficiency of the system.** Eff = (Pout / Pin) * 100% Pout = (1500 kg)(9.8 m/s²)(5.0 m) / 15 s Pout = 4900 W Eff = (4900 W / 12000 W) * 100% Eff = 41% **3. How much work must be done to stop a 2200 kg mass travelling at 33 m/s?** W = ΔKE = 1/2 * m * (vf² - vi²) W = 1/2 * (2200 kg) * (33 m/s)² W = 1.2 x 10^6 J **4. A carpenter holds a 21 kg sheet of plywood against a 2.5 m high ceiling for 35 seconds while his partner nails it in place. How much work did the carpenter do?** W = Fd W = 0 J **5. A workman on the CN Tower in Toronto dropped a 2.0 kg wrench. Given that the CN Tower is 300 m high, calculate the work done by the force of gravity on the wrench.** W = mgh = (2 kg)(9.81 m/s²)(300 m) W = 5.88 x 10³ J **6. A job is done slowly, and an identical job is done quickly. Both jobs require the same amount of work, but different amounts of what?** Power **7. A gardener exerts a 100 N force on a lawn mower handle at a 45° angle to the horizontal. How much work is done to move it 30 m?** F (100 N) / / 45° / / /________________ 30 m W = Fd cosθ W = (100 N)(30 m) cos(45°) W ≈ 2.1 x 10³ J **8. A pendulum with a bob of mass 0.750 kg is initially displaced to the left as shown.** A / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ B--------------C--------------D 0.250 m 0.240 m 0.661 m 0.650 m (a) How much heat energy is produced because of friction? EpA = mghA EpB = mghB H = EpA - EpB H = 0.75 kg * 9.81 m/s² * (0.35 m - 0.20 m) H ≈ 7.36 x 10^-2 J (b) If there were no friction, what would the speed of the bob be as it first passed through its lowest point? Ep = Ek mgh = 1/2 mv² v = √(2gh) = √(2 * 9.81 m/s² * 0.25 m) v ≈ 2.21 m/s