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Question Number 28905 by tawa tawa last updated on 01/Feb/18

A body rolls down a slope from a height of 100m. the velocity at the foot of the slope is 20 m/s. What percentage of the P.E is converted in K.E ? Answer: 20%

$$\mathrm{A}\:\mathrm{body}\:\mathrm{rolls}\:\mathrm{down}\:\mathrm{a}\:\mathrm{slope}\:\mathrm{from}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\:\mathrm{100m}.\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{at}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{slope}\:\mathrm{is}\:\:\mathrm{20}\:\mathrm{m}/\mathrm{s}.\:\:\mathrm{What}\:\mathrm{percentage}\:\mathrm{of}\:\mathrm{the}\:\boldsymbol{\mathrm{P}}.\boldsymbol{\mathrm{E}}\:\:\mathrm{is}\:\mathrm{converted}\:\mathrm{in}\:\:\boldsymbol{\mathrm{K}}.\boldsymbol{\mathrm{E}}\:\:? \\ $$$$ \\ $$$$\boldsymbol{\mathrm{A}}\mathrm{nswer}:\:\:\:\:\:\mathrm{20\%} \\ $$

Answered by mrW2 last updated on 01/Feb/18

PE=mgh KE=(1/2)mv^2 +(1/2)I(v^2 /R^2 )=(m/2)(1+(I/(mR^2 )))v^2 α=((KE)/(PE))=(v^2 /(2gh))(1+(I/(mR^2 ))) For a solid ball: I=(2/5)mR^2 ⇒α=(v^2 /(2gh))(1+(2/5))=((20^2 )/(2×10×100))×1.4=0.2×1.4=0.28=28% For a hollow ball: I=(2/3)mR^2 ⇒α=(v^2 /(2gh))(1+(2/3))=0.2×1.67=0.34=34% For a disc (cylinder): I=(1/2)mR^2 ⇒α=(v^2 /(2gh))(1+(1/2))=0.2×1.5=0.30=30% For a sliding object (without rotation): I=0 ⇒α=(v^2 /(2gh))(1+0)=0.2×1.0=0.20=20%

$${PE}={mgh} \\ $$$${KE}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{I}\frac{{v}^{\mathrm{2}} }{{R}^{\mathrm{2}} }=\frac{{m}}{\mathrm{2}}\left(\mathrm{1}+\frac{{I}}{{mR}^{\mathrm{2}} }\right){v}^{\mathrm{2}} \\ $$$$\alpha=\frac{{KE}}{{PE}}=\frac{{v}^{\mathrm{2}} }{\mathrm{2}{gh}}\left(\mathrm{1}+\frac{{I}}{{mR}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${For}\:{a}\:{solid}\:{ball}:\:{I}=\frac{\mathrm{2}}{\mathrm{5}}{mR}^{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\frac{{v}^{\mathrm{2}} }{\mathrm{2}{gh}}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{5}}\right)=\frac{\mathrm{20}^{\mathrm{2}} }{\mathrm{2}×\mathrm{10}×\mathrm{100}}×\mathrm{1}.\mathrm{4}=\mathrm{0}.\mathrm{2}×\mathrm{1}.\mathrm{4}=\mathrm{0}.\mathrm{28}=\mathrm{28\%} \\ $$$$ \\ $$$${For}\:{a}\:{hollow}\:{ball}:\:{I}=\frac{\mathrm{2}}{\mathrm{3}}{mR}^{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\frac{{v}^{\mathrm{2}} }{\mathrm{2}{gh}}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{0}.\mathrm{2}×\mathrm{1}.\mathrm{67}=\mathrm{0}.\mathrm{34}=\mathrm{34\%} \\ $$$$ \\ $$$${For}\:{a}\:{disc}\:\left({cylinder}\right):\:{I}=\frac{\mathrm{1}}{\mathrm{2}}{mR}^{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\frac{{v}^{\mathrm{2}} }{\mathrm{2}{gh}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0}.\mathrm{2}×\mathrm{1}.\mathrm{5}=\mathrm{0}.\mathrm{30}=\mathrm{30\%} \\ $$$$ \\ $$$${For}\:{a}\:{sliding}\:{object}\:\left({without}\:{rotation}\right):\:{I}=\mathrm{0} \\ $$$$\Rightarrow\alpha=\frac{{v}^{\mathrm{2}} }{\mathrm{2}{gh}}\left(\mathrm{1}+\mathrm{0}\right)=\mathrm{0}.\mathrm{2}×\mathrm{1}.\mathrm{0}=\mathrm{0}.\mathrm{20}=\mathrm{20\%} \\ $$

Commented by tawa tawa last updated on 01/Feb/18

God bless you sir. i really appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by NECx last updated on 01/Feb/18

This is Wonderful.

$$\mathrm{T}{his}\:{is}\:\mathbb{W}\boldsymbol{{onderful}}. \\ $$

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