cw7 - the mass oF the lighter blocks is 9 kg and the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
patel (kp9583) – CW7 – Mackie – (20211) 1 This print-out should have 2 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. DEADLINE IS CENTRAL TIME 001 10.0 points Joe pushes down the length oF the handle oF a 10 kg lawn spreader. The handle makes an angle oF 42 . 6 with the horizontal. Joe wishes to accelerate the spreader From rest to 1 . 31 m / s in 1 . 5 s. What Force must Joe apply to the handle? Correct answer: 11 . 8644 N. Explanation: The horizontal component oF the Force is F h = F cos θ. Let v be the fnal velocity oF the spreader. According to Newton’s second law, F h = m a h so F cos θ = m Δ v h Δ t F = m Δ v h Δ t cos θ = m v t cos θ = (10 kg) (1 . 31 m / s) (1 . 5 s) cos 42 . 6 = 11 . 8644 N keywords: 002 10.0 points The mass oF the heavier block is 18 kg and
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the mass oF the lighter blocks is 9 kg and the magnitude oF the Force oF the connecting string on the smaller block is 27 N. Assume: g = 9 . 8 m / s 2 and the horizontal surFace on which the objects slide is Friction-less. 9 kg 18 kg 27 N F Determine the Force F . Correct answer: 81 N. Explanation: Let : M = 9 kg , 2 M = 18 kg , and T = 27 N . M 2 M T T F ±or each body, F net = m a . ±or the leFtmost block, the tension acts to the right, so M a = T a = T M ±or the rightmost block, the Force acts to the right and the tension to the leFt, so 2 M a = F-T F = T + 2 M a = T + 2 M T M = 3 T = 3 (27 N) = 81 N ....
View Full Document

This note was uploaded on 12/10/2009 for the course PHYSICS 2021 taught by Professor Mihalisin during the Fall '09 term at Temple.

Ask a homework question - tutors are online