Question

Problem Set 2 (15 pts./ea) onstant of 6 N/m has a mass of 5kg attached weakly to it. A spring having a spring c The spring is compressed by 4m. a). What is the speed of the mass when the spring is released? b).The floor the mass is sliding on has a coefficient of kinetic friction of 1. s to for the mass to come to a stop?

Answer 1

Given Spring constant, $k=6N/m$

(a)When the compressed spring is released the elastic potential energy of spring is converted to kinetic energy of mass

$1/2mv^{2}=1/2kx^{2}$

$v=\sqrt{kx^{2}/m}$

$v=\sqrt{6N/m*(4m)^{2}/5kg}$

$v=\sqrt{6*(4)^{2}/5}$

ANSWER: $v=4.382m/s$

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(b)The Kinetic energy is lost due to friction while sliding on the floor

$KE=W_{Friction}$

$KE=F_{Friction}*d$

$KE=\mu N*d$

$1/2mv^{2}=\mu mg*d$

$v^{2}=2*\mu g*d$

$(4.382m/s)^{2}=2*0.11*9.81*d$

$19.2=2.1582*d$

$d=8.89263$

ANSWER: $d=8.9m$

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(c)initial velocity, $u=4.382m/s$

final velocity , $v=0m/s$

distance, $d=8.9m$

Use formula $v^{2}-u^{2}=2as$ to find deceleration

$0^{2}-(4.382m/s)^{2}=2a*8.9m$

$-(4.382)^{2}=17.8a$

$a=-1.0788m/s^{2}$

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Use Formula $v=u+at$ to find time

$0m/s=4.382m/s+(-1.0788)t$

$t=4.382/1.0788$

ANSWER: $t=4.062s$

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