Unit Test, Part 2
Answer the questions, using complete sentences. When you have finished, submit this assignment to your teacher by the due date for full credit. (8 points)
0. A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact southeast with a force of 2.00 × 104 N. The sailboat travels the distance in 1.0 h. How much work was done by the wind? What was the wind’s power? Your response should include all of your work and a free-body diagram. Answer:
Power is work done per unit time and
workdone is force times the displacement.
Hence power equal force * displacement / time.
P = 2e4*10 / 3600s = 55.56 W.
0. A sailor pushes a 100.0 kg crate up a ramp that is 3.00 m high and 5.00 m long onto the deck of a ship. He exerts a 650.0 N force parallel to the ramp. What is the mechanical advantage of the ramp? What is the efficiency of the ramp? Your response should include all of your work and a free-body diagram. Answer:
650 N is mvoed through a distance of 5m.
The result is
100*9.8 N is moved up thrugh a distance of 3m.
Mechnical advanatage is 100*9.8 / 650 = 1.5.
The maximum mechanical advanatage can be 5m / 3 m = 1.66
Efficiency = 1.5 / 1.66 = 0.90 = 90%
0. A man lifts various loads with the same lever. The distance of the applied force from the fulcrum is 2.00 m and the distance from the fulcrum to the load is 0.500 m. A graph of resistance force vs. effort force is shown. What is the mechanical advantage of the lever? What is the ideal mechanical advantage of the lever? What is the efficiency of the lever? Show your work.
mechanical energy is always constant
KE+ PE= constant
Mechanical advantage = 0.25
Ideal mechanical advantage = 1
Efficiency of the lever = 0.25 or 25%
work: Therefore AMA = (F(0.500m)) / (F(2.00m)) = 0.25
Since AMA = 0.25 , and IMA = 1
% Eff = AMA / IMA = 0.25 / 1.00 = 0.25 = 25% Efficient
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