ommon/Take TutorialAssignment aspx Date: 3/28/2018 11590 00 PM End Date: 3/28/20

Question

ommon/Take TutorialAssignment aspx Date: 3/28/2018 11590 00 PM End Date: 3/28/2018 11:59.00 PM Homework 6 Begin Date:39/2018 120000 AM-Due (6%) Problem 14: A block ofmass 4.6kg is sattang on a frictionless ramp with a pring at the bottions that has a spring constant of 480 Nim (refer to the figure). The angle of the ramp with respect to the horizontal is 399 Ctheesperita.com How fr nete meen irthr UR33% Part (a) The block, start nghcen rest slides down the ramp spring compressed as the block comes to momentary rest? moving rnight aftes it comes off the spring? v2.46491 ance 58 an before hittne the speng in meters per second, is tdhe block 33% Part (b) After the block comes to rest the spr ng pubnthe ? ck back up the ran p How fat Grade Semmry ed ed per atenpe atan acotan sinho ted ted rted Degrees O Radians CLEAR eted leted al potential e erty. joules, between the ong nal position of th" block at the top of the 33% Part (c) What is the change ofthpastato ramp and the position of the block when the spring is fully compeessed

Explanation / Answer

here,

mass , m = 4.6 kg

spring constant , K =480 N/m

theta = 39 degree

a)

s = 58 cm = 0.58 m

let the spring compresses x m

using conservation of energy

0.5 * K * x^2 = m * g * ( s +x) * sin(theta)

0.5 * 480 * x^2 = 4.6 * 9.81 * (0.58 + x ) * sin(39)

solving for x

x = 0.33 m

the spring compresses is 32.8 cm

b)

let the final speed of box be v

using conservation of energy

0.5 * k * x^2 = m * g * x * sin(theta) + 0.5 * m * v^2

0.5 * 480 * 0.33^2 = 4.6 * 9.81 * 0.33 * sin(39) + 0.5 * 4.6 * v^2

solving for v

v = 2.7 m/s

the final speed of block is 2.7 m/s

c)

the change in potential energy , PE = m * g * ( s + x) * sin(theta)

PE = 4.6 * 9.81 * ( 0.58 + 0.33) * sin(39) J

PE = 25.8 J

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