2. A force applied parallel to an inclined plane is used to raise a load of \( 500 \mathrm{~N} \) through a height of \( 3.0 \mathrm{~m} \). The effort moved through a distance of \( 30 \mathrm{~m} \) along the plane. i) How much work is done on the load? ii) If the efficiency of the inclined plane is \( 80 \% \), determine the effort applied. iii) How much work is done by the effort?
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0 m. So, Work = 500 N x 3.0 m = 1500 J Show more…
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The inclined plane depicted in long and rises $3.0$ m. ( $a$ ) What minimum force $F$ parallel to the plane is required to slide a 20 -kg box up the plane if friction is neglected? ( $b$ ) What is the IMA of the plane? ( $c$ ) Find the AMA and efficiency if a $64-\mathrm{N}$ force is actually required. (a) There are several ways to approach this. Let's consider energy. Since there is no friction, the work done by the pushing force, (F) $(15 \mathrm{~m})$, must equal the lifting work done, $(20 \mathrm{~kg})(9.81$ $\mathrm{m} / \mathrm{s}^{2}$ ) $(3.0 \mathrm{~m})$. Equating these two expressions and solving for $\mathrm{F}$ gives $F=39 \mathrm{~N}$ $(b)$ $\mathrm{IMA}=\frac{\text { Distance moved by } F}{\text { Distance } F_{w} \text { is lifted }}=\frac{15 \mathrm{~m}}{3.0 \mathrm{~m}}=5.0$ (c) $\mathrm{AMA}=$ Force ratio $=\frac{F_{W}}{F}=\frac{196 \mathrm{~N}}{64 \mathrm{~N}}=3.06=3.1$ Efficiency $=\frac{A M A}{I M A}=\frac{3.06}{5.0}=0.61=61 \%$ $\mathrm{O} r$, as a check. Efficiency $=\frac{\text { Work output }}{\text { Work input }}=\frac{\left(F_{W}\right)(3.0 \mathrm{~m})}{(F)(15 \mathrm{~m})}=0.61=61 \%$
An inclined plane is $10.0 \mathrm{~m}$ long and $2.50 \mathrm{~m}$ high. (a) Find its mechanical advantage. (b) A resistance of $727 \mathrm{~N}$ is pushed up the plane. What effort is needed? (c) An effort of $20 \overline{0} \mathrm{~N}$ is applied to push an $815-\mathrm{N}$ resistance up the inclined plane. Is the effort enough?
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A plane designed for vertical takeoff has a mass of $8.0 \times 10^{3} \mathrm{kg}$. Find the net work done by all forces on the plane as it accelerates upward at $1.0 \mathrm{m} / \mathrm{s}^{2}$ through a distance of $30.0 \mathrm{m}$ after starting from rest. (See Sample Problem 5A.)
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