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Question Number 49559 by Rio Michael last updated on 07/Dec/18
Commented by Rio Michael last updated on 07/Dec/18
The figure below shows an object of mass 60kg pulled up by an effort of 4000J through an inclined plane 30° to the horizontal. a) Sketch the diagram showing all forces acting on the object as it moves up the inclined plane. b) The object is allowed to fall through a vertical height of 6m calculate the change in potential energy. c) Calculate the efficiency of the system above hence state why its less then 100%.
$${The}\:{figure}\:{below}\:{shows}\:{an}\:{object}\:{of}\:{mass}\:\mathrm{60}{kg}\:{pulled}\:{up} \\ $$$${by}\:{an}\:{effort}\:{of}\:\mathrm{4000}{J}\:{through}\:{an}\:{inclined}\:{plane}\:\mathrm{30}°\:{to}\:{the}\: \\ $$$${horizontal}. \\ $$$$\left.{a}\right)\:{Sketch}\:{the}\:{diagram}\:{showing}\:{all}\:{forces}\:{acting}\:{on}\:{the}\:{object} \\ $$$${as}\:{it}\:{moves}\:{up}\:{the}\:{inclined}\:{plane}. \\ $$$$\left.{b}\right)\:{The}\:{object}\:{is}\:{allowed}\:{to}\:{fall}\:{through}\:{a}\:{vertical}\:{height}\:{of}\:\mathrm{6}{m} \\ $$$${calculate}\:{the}\:{change}\:{in}\:{potential}\:{energy}. \\ $$$$\left.{c}\right)\:{Calculate}\:{the}\:{efficiency}\:{of}\:{the}\:{system}\:{above}\:{hence}\:{state} \\ $$$${why}\:{its}\:{less}\:{then}\:\mathrm{100\%}. \\ $$
Answered by afachri last updated on 07/Dec/18
Commented by afachri last updated on 08/Dec/18
sir, pardon me. i did a mistake in the diagram. the arrow of w cos30 should be w , and w arrow should be w cos30. i am so sorry.
$$\mathrm{sir},\:\mathrm{pardon}\:\mathrm{me}.\:\mathrm{i}\:\mathrm{did}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{diagram}.\:\mathrm{the}\:\mathrm{arrow}\:\mathrm{of}\:{w}\:{cos}\mathrm{30}\: \\ $$$$\mathrm{should}\:\mathrm{be}\:{w}\:,\:\mathrm{and}\:{w}\:\mathrm{arrow}\:\:\mathrm{should}\:\mathrm{be}\: \\ $$$${w}\:{cos}\mathrm{30}.\:\mathrm{i}\:\mathrm{am}\:\mathrm{so}\:\mathrm{sorry}. \\ $$$$ \\ $$
Commented by Rio Michael last updated on 08/Dec/18
no problem sir
$${no}\:{problem}\:{sir} \\ $$
Commented by Rio Michael last updated on 08/Dec/18
thanks sir i saw that already
$${thanks}\:{sir}\:{i}\:{saw}\:{that}\:{already} \\ $$
Commented by afachri last updated on 07/Dec/18
your welcome Sir. the other answer below.
$$\mathrm{your}\:\mathrm{welcome}\:\mathrm{Sir}.\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{answer}\:\mathrm{below}. \\ $$
Answered by afachri last updated on 07/Dec/18
B) let′s say the gravity conts : 10 m/s^2 △Potential energy = m.g.△h = (60 Kg)(10 m/s^2 )(6 m) = 3600 J C) find first : w sin(30) and F w sin(30°) = m × g × sin(30°) = 60 × 10 (1/(2 )) = 300 J and, F = (W/s) = (4000 J) / 12 m = ((1000)/3) N so, the efficiency of system will be : e = ((w sin (30°))/F) × 100% = 90% ⇒ reason : cause applied effort (4000 J ; 300 N) is over than the force needed (((1000)/3) N)
$$\left.\boldsymbol{\mathrm{B}}\right)\:\mathrm{let}'\mathrm{s}\:\:\mathrm{say}\:\mathrm{the}\:\mathrm{gravity}\:\mathrm{conts}\::\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\bigtriangleup\boldsymbol{{Potential}}\:\boldsymbol{{energy}}\:\:=\:\:\boldsymbol{{m}}.\boldsymbol{{g}}.\bigtriangleup\boldsymbol{{h}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\left(\mathrm{60}\:\mathrm{Kg}\right)\left(\mathrm{10}\:\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right)\left(\mathrm{6}\:\mathrm{m}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{3600}\:\mathrm{J} \\ $$$$\left.\boldsymbol{\mathrm{C}}\right)\:\:\mathrm{find}\:\mathrm{first}\::\:\:\:{w}\:{sin}\left(\mathrm{30}\right)\:\mathrm{and}\:{F} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{w}\:{sin}\left(\mathrm{30}°\right)\:\:=\:\:{m}\:×\:{g}\:×\:{sin}\left(\mathrm{30}°\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{60}\:\:×\:\:\mathrm{10}\:\:\frac{\mathrm{1}}{\mathrm{2}\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{300}\:\mathrm{J} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{and}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{F}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\:\frac{{W}}{{s}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{4000}\:\boldsymbol{\mathrm{J}}\right)\:/\:\mathrm{12}\:\mathrm{m}\:=\:\frac{\mathrm{1000}}{\mathrm{3}}\:\mathrm{N} \\ $$$$\:\:\:\:\:\:\:\mathrm{so},\:\mathrm{the}\:\mathrm{efficiency}\:\mathrm{of}\:\mathrm{system}\:\mathrm{will}\:\mathrm{be}\:: \\ $$$$\:\:\:\:\:\:{e}\:\:=\:\:\frac{{w}\:{sin}\:\left(\mathrm{30}°\right)}{{F}}\:×\:\mathrm{100\%} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{90\%}\:\:\Rightarrow\:\:\mathrm{reason}\::\:\mathrm{cause}\:\mathrm{applied}\:\mathrm{effort}\:\left(\mathrm{4000}\:\mathrm{J}\:;\:\mathrm{300}\:\mathrm{N}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{is}\:\mathrm{over}\:\mathrm{than}\:\mathrm{the}\:\mathrm{force}\:\mathrm{needed}\:\left(\frac{\mathrm{1000}}{\mathrm{3}}\:\mathrm{N}\right) \\ $$$$ \\ $$$$ \\ $$

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