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    <title>Recent Posts in 'Physics - Work done' | sgForums.com</title>
    <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
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      <title>Physics - Work done replied by secretliker @ Sun, 13 Jul 2008 02:17:29 +0800</title>
      <description>&lt;p&gt;Yup. It's 0.8J (I carelessly made it 80J).&lt;/p&gt;</description>
      <pubDate>Sun, 13 Jul 2008 02:17:29 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8237250</guid>
      <author>secretliker</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by arigatoast @ Sat, 12 Jul 2008 12:57:54 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by bonkysleuth:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;Eh, why is it that on the top PE = KE? I don't think I've ever
heard of that before. Can anyone explain?&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;I only know a scenario when there is G.P.E on top but no K.E and
vice versa for the bottom.&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;oops sorry.. write wrongly.&lt;/p&gt;
&lt;p&gt;on top of the ramp.. just b4 it slide down Max PE with 0.3J KE
as stated in question.&lt;/p&gt;
&lt;p&gt;i think u mean lose 0.5 J of GPE instead of 0.5 of GPE. so lost
in GPE = gain in KE=0.5J&lt;/p&gt;
&lt;p&gt;0.5J+0.3J = 0.8J (Ans)&lt;/p&gt;
&lt;p&gt;ignore my previous post. thanks.&lt;/p&gt;</description>
      <pubDate>Sat, 12 Jul 2008 12:57:54 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8236044</guid>
      <author>arigatoast</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by arigatoast @ Fri, 11 Jul 2008 22:59:23 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by secretliker:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;At top of ramp, car has KE (0.30J) and GPE (this is equal to 50J
since the car loses 50J of GPE when it travels to the bottom).&lt;/p&gt;
&lt;p&gt;Assuming no energy is lost as thermal energy, all GPE is
converted to KE.&lt;/p&gt;
&lt;p&gt;Thus KE at bottom of ramp = 80J?&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;Hi secret liker.. how come is 80J? the energy mentioned are KE
0.3J and 0.5 of GPE lost. U mean 0.8 J?&lt;/p&gt;</description>
      <pubDate>Fri, 11 Jul 2008 22:59:23 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8235204</guid>
      <author>arigatoast</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by eagle @ Fri, 11 Jul 2008 08:45:27 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by Uncertain:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;&lt;br /&gt;
Ok I think eagle will kill me if i keep on taking his lobang...
hehe BUT he is the best in physics here...&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Nevertheless, these are my thoughts...&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;(i)&amp;nbsp; 1) Energy is needed to overcome the inertia of the
wheel so energy is wasted.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 2) Like what u said.... energy is lost
in form of sound, heat and so on energy.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;ii) is 0.50J of PE or 1/2 of PE loss? Be accurate in ur
question.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;if 0.50J, then follow secretliker&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;if 1/2, then follow agriatoast.&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;Dun worry, just take&lt;/p&gt;
&lt;p&gt;I quite busy, and a bit sick...&lt;/p&gt;
&lt;p&gt;In homework forum, as long as you can do the question and
explain, just go ahead! We are sharing afterall, no difference&lt;/p&gt;</description>
      <pubDate>Fri, 11 Jul 2008 08:45:27 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8233552</guid>
      <author>eagle</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by Uncertain @ Fri, 11 Jul 2008 08:40:07 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by bonkysleuth:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;&lt;br /&gt;
thanks so much. i was thinking about that too! by the way, can you
tell me why a hydroelectric generator device wastes energy? the
question goes something like "the water supplies 2000J of energy
every second to the wheel. the electrical energy output is 1200 J
every second.&lt;/p&gt;
&lt;p&gt;1 way which it wastes energy is perhaps through the loss of
thermal energy i suppose. during the conversions of different forms
of energy, there is bound to be some energy which is lost as heat
energy. what is the second reason? i can't think of another.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;and there;s this question. if is of a car moving on a horizontal
section until it reaches a ramp and moves downwards. length of the
ramp is 0.9 m .&lt;/p&gt;
&lt;p&gt;(ii) the car has 0.30 J of kinetic energy at the topof the ramp
and loses 0.50 of potential energy as it moves to the bottom of the
ramp. calculate kinetic energy of car at bottom of the ramp.&lt;/p&gt;
&lt;p&gt;i tried to equate using principle of conservation of energy.&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;&lt;br /&gt;
Ok I think eagle will kill me if i keep on taking his lobang...
hehe BUT he is the best in physics here...&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Nevertheless, these are my thoughts...&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;(i)&amp;nbsp; 1) Energy is needed to overcome the inertia of the
wheel so energy is wasted.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 2) Like what u said.... energy is lost
in form of sound, heat and so on energy.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;ii) is 0.50J of PE or 1/2 of PE loss? Be accurate in ur
question.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;if 0.50J, then follow secretliker&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;if 1/2, then follow agriatoast.&lt;/p&gt;</description>
      <pubDate>Fri, 11 Jul 2008 08:40:07 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8233550</guid>
      <author>Uncertain</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by secretliker @ Fri, 11 Jul 2008 00:48:34 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by bonkysleuth:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;and there;s this question. if is of a car moving on a horizontal
section until it reaches a ramp and moves downwards. length of the
ramp is 0.9 m .&lt;/p&gt;
&lt;p&gt;(ii) the car has 0.30 J of kinetic energy at the topof the ramp
and loses 0.50 of potential energy as it moves to the bottom of the
ramp. calculate kinetic energy of car at bottom of the ramp.&lt;/p&gt;
&lt;p&gt;i tried to equate using principle of conservation of energy.&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;At top of ramp, car has KE (0.30J) and GPE (this is equal to 50J
since the car loses 50J of GPE when it travels to the bottom).&lt;/p&gt;
&lt;p&gt;Assuming no energy is lost as thermal energy, all GPE is
converted to KE.&lt;/p&gt;
&lt;p&gt;Thus KE at bottom of ramp = 80J?&lt;/p&gt;</description>
      <pubDate>Fri, 11 Jul 2008 00:48:34 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8233260</guid>
      <author>secretliker</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by bonkysleuth @ Fri, 11 Jul 2008 00:24:43 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by arigatoast:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;wad i think is&lt;/p&gt;
&lt;p&gt;1) energy required to overcome the inertia of the wheel as the
wheel is likely to be large.&lt;/p&gt;
&lt;p&gt;2)friction between water molecules and the wheel.&lt;/p&gt;
&lt;p&gt;&lt;img src="/images/emoticons/kde-3.5.8/set1/wink.png" alt=
"wink.png" /&gt;&lt;/p&gt;
&lt;p&gt;2nd question... 0.3 x 0.5= 0.15J= KE at bottom (ans)&lt;/p&gt;
&lt;p&gt;on top, PE = KE = 0.3&lt;/p&gt;
&lt;p&gt;as u go down to bottom lose 1/2 of it.&lt;/p&gt;
&lt;p&gt;at bottom, energy converted to KE again.&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;Eh, why is it that on the top PE = KE? I don't think I've ever
heard of that before. Can anyone explain?&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;I only know a scenario when there is G.P.E on top but no K.E and
vice versa for the bottom.&lt;/p&gt;</description>
      <pubDate>Fri, 11 Jul 2008 00:24:43 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8233222</guid>
      <author>bonkysleuth</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by arigatoast @ Thu, 10 Jul 2008 22:58:09 +0800</title>
      <description>&lt;p&gt;wad i think is&lt;/p&gt;
&lt;p&gt;1) energy required to overcome the inertia of the wheel as the
wheel is likely to be large.&lt;/p&gt;
&lt;p&gt;2)friction between water molecules and the wheel.&lt;/p&gt;
&lt;p&gt;&lt;img src="/images/emoticons/kde-3.5.8/set1/wink.png" alt=
"wink.png" /&gt;&lt;/p&gt;
&lt;p&gt;2nd question... 0.3 x 0.5= 0.15J= KE at bottom (ans)&lt;/p&gt;
&lt;p&gt;on top, PE = KE = 0.3&lt;/p&gt;
&lt;p&gt;as u go down to bottom lose 1/2 of it.&lt;/p&gt;
&lt;p&gt;at bottom, energy converted to KE again.&lt;/p&gt;</description>
      <pubDate>Thu, 10 Jul 2008 22:58:09 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8233073</guid>
      <author>arigatoast</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by bonkysleuth @ Thu, 10 Jul 2008 22:24:04 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by Uncertain:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;Work done against friction = energy lose due to friction&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;So work done against friction = 100 x 3 = 300J&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;&lt;br /&gt;
thanks so much. i was thinking about that too! by the way, can you
tell me why a hydroelectric generator device wastes energy? the
question goes something like "the water supplies 2000J of energy
every second to the wheel. the electrical energy output is 1200 J
every second.&lt;/p&gt;
&lt;p&gt;1 way which it wastes energy is perhaps through the loss of
thermal energy i suppose. during the conversions of different forms
of energy, there is bound to be some energy which is lost as heat
energy. what is the second reason? i can't think of another.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;and there;s this question. if is of a car moving on a horizontal
section until it reaches a ramp and moves downwards. length of the
ramp is 0.9 m .&lt;/p&gt;
&lt;p&gt;(ii) the car has 0.30 J of kinetic energy at the topof the ramp
and loses 0.50 of potential energy as it moves to the bottom of the
ramp. calculate kinetic energy of car at bottom of the ramp.&lt;/p&gt;
&lt;p&gt;i tried to equate using principle of conservation of energy.&lt;/p&gt;</description>
      <pubDate>Thu, 10 Jul 2008 22:24:04 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8233021</guid>
      <author>bonkysleuth</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by Uncertain @ Thu, 10 Jul 2008 21:14:56 +0800</title>
      <description>&lt;p&gt;Work done against friction = energy lose due to friction&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;So work done against friction = 100 x 3 = 300J&lt;/p&gt;</description>
      <pubDate>Thu, 10 Jul 2008 21:14:56 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8232875</guid>
      <author>Uncertain</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
    </item>
    <item>
      <title>Physics - Work done replied by bonkysleuth @ Thu, 10 Jul 2008 20:50:41 +0800</title>
      <description>&lt;p&gt;A person exerts a horizontal force of 500 N on a box, which also
experiences a friction force of 100 N. How much work is done
against friction when the box moves a horizontal direction of 3
m?&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Do you get the resultant force and multiply by 3 m? 400 X 3 =
1200? I don't really get the question when it says "work done
against friction".&lt;/p&gt;</description>
      <pubDate>Thu, 10 Jul 2008 20:50:41 +0800</pubDate>
      <guid isPermaLink="false">politics.sgforums.com:2297:323772:8232819</guid>
      <author>bonkysleuth</author>
      <link>http://politics.sgforums.com/forums/2297/topics/323772</link>
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