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Unformatted text preview: homework 04 – GADHIA, TEJAS – Due: Feb 14 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Despite a very strong wind, a tennis player manages to hit a tennis ball with her racquet so that the ball passes over the net and lands in her opponents court. Consider the following forces: 1  A downward force of gravity. 2  A force by the ”hit”. 3  A force exerted by the air. Which of the above forces is (are) acting on the tennis ball after it has left contact with the racquet and before it touches the ground? 1. 1 and 2 2. 2 and 3 3. 1 and 3 correct 4. 1 only 5. 1, 2 and 3 Explanation: After the contact with the racquet, only the air resistance and the gravity act upon the ball. Question 2 part 1 of 1 10 points An elevator is being lifted up an el evator shaft at a constant speed by a steel cable as shown in the figure be low. All frictional effects are negligible. steel cable Elevator going up at constant speed In this situation, forces on the elevator are such that 1. the upward force by the cable is greater than the sum of the downward force of gravity and a downward force due to the air. 2. the upward force by the cable is equal to the downward force of gravity. correct 3. the upward force by the cable is greater than the downward force of gravity. 4. None of these. (The elevator goes up because the cable is being shortened, not be cause an upward force is exerted on the eleva tor by the cable.) 5. the upward force by the cable is smaller than the downward force of gravity. Explanation: Since the elevator is being lifted at a con stant speed, the net force on it is zero, there fore, the upward force by the cable is equal to the downward force of gravity. Question 3 part 1 of 1 10 points A 1 ton truck provides an acceleration of 4 . 6 ft / s 2 to a 8 ton trailer. If the truck exerts the same force on the road while pulling a 23 . 5 ton trailer, what acceleration results? Correct answer: 1 . 6898 ft / s 2 (tolerance ± 1 homework 04 – GADHIA, TEJAS – Due: Feb 14 2007, 4:00 am 2 %). Explanation: When the truck is pulling the 8 ton trailer, it is imparting a force on its own mass and the trailer’s mass, (1 ton + 8 ton). This gives the following force that the road exerts on the truck: F = ( m truck + m trailer 1 ) a 1 = (1 ton + 8 ton) (4 . 6 ft / s 2 ) = 41 . 4 ton · ft / s 2 This is also the magnitude of the force that the truck exerts on the road. Assuming the same force is exerted on the road, and the truck now pulls a trailer of mass 23 . 5 ton, the following equation must be satisfied: F = ( m truck + m trailer 2 ) a 2 Solving for the acceleration, a 2 , gives a 2 = F m truck + m trailer 2 = 41 . 4 ton · ft / s 2 1 ton + 23 . 5 ton = 1 . 6898 ft / s 2 Question 4 part 1 of 3 10 points A dragster and driver together have mass 888 . 1 kg . The dragster, starting from rest, attains a speed of 26 . 8 m / s in 0 . 65 s ....
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 Spring '08
 Turner
 Force, Friction, Mass, Work, General Relativity, Correct Answer

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