# PHYSICS - originally compressed from its equilibrium...

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E XAMPLES 1) A 17.1 kg block is dragged over a rough, hor- izontal surface by a constant force of 147 N acting at an angle of 31◦ above the horizon- tal. The block is displaced 20.4 m, and the coefficient of kinetic friction is 0.218. Find the work done by the 147 N force. The acceleration of gravity is 9.8 m/s2 . Answer in units of J. ANSWER : W F = F d cosq F = (112 N)(15.5 m)cos(40.0 o ) = 1330J = (147)(20.4)(cos(31))= 2570.473 2) Two masses of 25 kg and 17 kg are suspended by a pulley that has a radius of 4.1 cm and a mass of 8 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 2 m apart. Determine the speeds of the two masses as they pass each other. Treat the pulley as a uniform disk. The acceleration of gravity is 9.8 m/s2 . Answer in units of m/s. ANSWER : (m1 - m2)g = a*(m1 + m2 + M/2) =(25-17)*9.8 = a(25+27+[8/2]) a=1.7043 v^2 = vo^2 + 2aS v^2 = 0 + 2*1.7043*(2/2) =1.846 3) 002 (part 2 of 2) 0 points If M is 5 kg and the spring between the masses has a spring constant of 6300 N/m, how much was the spring

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Unformatted text preview: originally compressed from its equilibrium length? Correct answer: 0.27618 m. 4) A child bounces a 47 g superball on the side- walk. The velocity change of the superball is from 22 m/s downward to 12 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Answer in units of N. ANSWER : (0.047)*(22+12)=1.598 =1.598*800=1278.4 5) A 1 kg steel ball strikes a wall with a speed of 14.1 m/s at an angle of 38◦ with the normal to the wall. It bounces off with the same speed and angle as shown in the figure. If the ball is in contact with the wall for 0.143 s, what is the magnitude of the average force exerted on the ball by the wall? ANSWER : (1)(14.1)sin(52)=11.11 kg m/s 11.11 x 2 = 22.22 m/s = 22.22/0.143=155.398 6) m2) immediately after the collison and before the pendulum starts to swing upwards. Answer in units of m/s ANSWER : max height= 0.056 Max GPE= mgh= (795)(9.81)(.056)=436.7412 J ½(795)v 2 =436.7412 V=1.048...
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## This note was uploaded on 02/23/2012 for the course PHYS 2425 taught by Professor Sherman during the Summer '08 term at Collin College.

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PHYSICS - originally compressed from its equilibrium...

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