HW 7-solutions

HW 7-solutions - kuruvila (lk5992) HW 7 opyrchal (11113) 1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: kuruvila (lk5992) HW 7 opyrchal (11113) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A(n) 769 kg elevator starts from rest. It moves upward for 2 . 93 s with a constant acceleration until it reaches its cruising speed of 2 m / s. The acceleration of gravity is 9 . 8 m / s 2 . Find the average power delivered by the elevator motor during the period of this accel- eration. Correct answer: 8 . 06112 kW. Explanation: Let : m = 769 kg , t = 2 . 93 s , and v = 2 m / s , The height y is y = v ave t = 1 2 v t = 1 2 (2 m / s) (2 . 93 s) = 2 . 93 m . Since the elevator starts from rest, the power P supplied by the motor, totally transferred into Kinetic Energy K and Potential Energy U , where E = K + U , is P = E t = U + K t = mg y + 1 2 mv 2 t = mg 1 2 v t + 1 2 mv 2 t = mv 2 t ( g t + v ) = (769 kg) (2 m / s) 2 (2 . 93 s) [(9 . 8 m / s 2 ) (2 . 93 s) + 2 m / s] = 8061 . 11 W = 8 . 06112 kW . 002 10.0 points A block of mass 0 . 25 kg is placed on a verti- cal spring of constant 2188 N / m and pushed downward, compressing the spring 0 . 07 m. After the block is released it leaves the spring and continues to travel upward. The acceleration of gravity is 9 . 8 m / s 2 . What height above the point of release will the block reach if air resistance is negligible? Correct answer: 2 . 188 m. Explanation: Given : m = 0 . 25 kg , x = 0 . 07 m , k = 2188 N / m , and g = 9 . 8 m / s 2 . Choose U g = 0 at the level of the release point. K i = K f = 0 , so from conservation of en- ergy, ( U g + U s ) i = ( U g + U s ) f 0 + U s,i = U g,f + 0 1 2 k x 2 = mg h h = k x 2 2 mg = (2188 N / m)(0 . 07 m) 2 2(0 . 25 kg)(9 . 8 m / s 2 ) = 2 . 188 m . 003 (part 1 of 2) 10.0 points A pendulum consists of a sphere of mass 1 . 3 kg attached to a light cord of length 11 . 1 m as in the figure below. The sphere is released from rest when the cord makes a 48 . 7 angle with the vertical, and the pivot at P is frictionless.is frictionless....
View Full Document

Page1 / 5

HW 7-solutions - kuruvila (lk5992) HW 7 opyrchal (11113) 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online