(a) Explain the term work (b) Draw a diagram of a pulley system with a velocity ratio of 5. (c) A man pulls up a box of mass 70kg using an inclined plane of...

(a) Work

Work is done when a force moves a body through a distance in the direction of the force. Thus,

\[W=Fs\]

where \(F\) is the force and \(s\) is the displacement in its direction. Work is a scalar quantity and its SI unit is the joule (J).

(b) Pulley system of velocity ratio 5

There are five rope sections supporting the movable block. Hence,

\[V.R.=5\]

(c) Inclined plane

The weight of the box is

\[W=mg=70\times10=700\ \text{N}.\]

The plane makes an angle \(\theta\) with the horizontal such that

\[\sin\theta=\frac{h}{L}=\frac{2.5}{5}=0.5,\qquad \theta=30^\circ.\]

The force diagram for the box moving up the plane is shown below.

(i) Minimum effort applied

Since the box moves at uniform speed, the effort balances the component of weight down the plane and the frictional force:

\[E=mg\sin\theta+F=700(0.5)+100=450\ \text{N}.\]

(ii) Velocity ratio of the plane

\[V.R.=\frac{\text{distance moved by effort}}{\text{distance moved by load}}=\frac{L}{h}=\frac{5}{2.5}=2.\]

(iii) Mechanical advantage of the plane

\[M.A.=\frac{\text{Load}}{\text{Effort}}=\frac{700}{450}=1.56.\]

(iv) Efficiency of the plane

\[\eta=\frac{M.A.}{V.R.}\times100\%=\frac{1.56}{2}\times100\%=77.8\%\approx78\%.\]

(v) Energy lost in the system

Energy is lost in overcoming friction:

\[E_{\text{lost}}=F\times L=100\times5=500\ \text{J}.\]

(vi) Work output

\[W_{\text{output}}=mgh=70\times10\times2.5=1750\ \text{J}.\]

(vii) Total power developed by the man

The work input is

\[W_{\text{input}}=E\times L=450\times5=2250\ \text{J}.\]

Therefore,

\[P=\frac{W_{\text{input}}}{t}=\frac{2250}{50}=45\ \text{W}.\]