HW5 - homework 05 GAUTHIER, JOSEPH Due: Oct 5 2007, 1:00 am...

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Unformatted text preview: homework 05 GAUTHIER, JOSEPH Due: Oct 5 2007, 1:00 am 1 Question 1, chap 7, sect 1. part 1 of 2 10 points A 17 . 2 kg block is dragged over a rough, horizontal surface by a constant force of 105 N acting at an angle of 28 . 7 above the horizon- tal. The block is displaced 86 . 5 m, and the coefficient of kinetic friction is 0 . 198. The acceleration of gravity is 9 . 8 m / s 2 . 17 . 2 kg = 0 . 198 1 5 N 2 8 . 7 Find the work done by the 105 N force. Correct answer: 7966 . 68 J (tolerance 1 %). Explanation: Consider the force diagram F m g n f k Work is W = vector F vectors , where vectors is the distance traveled. In this problem vectors = 5 is only in the x direction. W F = F x s x = F s x cos = (105 N) (86 . 5 m) cos28 . 7 = 7966 . 68 J , . Question 2, chap 7, sect 1. part 2 of 2 10 points Find the magnitude of the work done by the force of friction. Correct answer: 2023 . 32 J (tolerance 1 %). Explanation: To find the frictional force, F friction = N , we need to find N from vertical force balance. Note that N is in the same direction as the y component of F and opposite the force of gravity.Thus F sin + N = m g so that N = m g- F sin . Thus the friction force is vector F friction =- N =- ( m g- F sin ) . The work done by friction is then W = vector F friction vectors =-| f || s | =- ( m g- f sin ) s x =- . 198[(17 . 2 kg) (9 . 8 m / s 2 )- (105 N) sin28 . 7 ] (86 . 5 m) =- 2023 . 32 J . Question 3, chap 7, sect 1. part 1 of 1 10 points If 2.2 J of work is done in raising a 175 g apple, how far is it lifted? Correct answer: 1 . 2828 m (tolerance 1 %). Explanation: Basic Concepts: W applied = F applied d cos = mgd since = 0 cos = 1. 1 J = 1 N m F = mg 1 N = 1 kg m / s 2 Given: W applied = 2 . 2 J m = 175 g g = 9 . 81 m / s 2 Solution: The force and displacement are parallel, so the distance is given by homework 05 GAUTHIER, JOSEPH Due: Oct 5 2007, 1:00 am 2 d = W applied mg = 2 . 2 J (175 g)(9 . 8 m / s 2 ) 1000 g 1 kg = 1 . 2828 m Question 4, chap 7, sect 1. part 1 of 1 10 points A tugboat exerts a constant force of 6840 N on a ship moving at constant speed through a harbor. How much work does the tugboat do on the ship if each moves a distance of 3 . 43 km? Correct answer: 23 . 4612 MJ (tolerance 1 %). Explanation: Given : F net = 6840 N and d = 3 . 43 km . The net force F net = F x = F . The force and the displacement are parallel, so the net work is W net = F net d = (6840 N) (3 . 43 km) parenleftbigg 1000 m 1 km parenrightbigg parenleftbigg 1 MJ 10 6 J parenrightbigg = 23 . 4612 MJ , since 1 J = 1 N m....
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HW5 - homework 05 GAUTHIER, JOSEPH Due: Oct 5 2007, 1:00 am...

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