Question

A screw jack of pitch $5$mm is used to lift the tyre of a vehicle of load \[200\]kg with the help of a handle of length \[0.5\]m. Neglecting the frictional force between screw and nut of the jack, the least force required to raise the load is:

\[\begin{align}
  & A.1.2N \\
 & B.2.2N \\
 & C.3.2N \\
 & D.4.2N \\
\end{align}\]

Answer 1

Hint: Define angular simple harmonic motion. Obtain the expression for the angular displacement of the particle. From the given quantities, find the time for the angular displacement. Obtain the expression for angular velocity by differentiating the angular displacement. By putting the given values, we can find the answer.

Formula used:
$\theta ={{\theta }_{A}}\sin \left( \omega t \right)$
$w=\dfrac{d\theta }{dt}$
$\omega =\dfrac{2\pi }{T}$


Complete step by step answer:
A screw jack is a type of jack, which can lift a moderately heavy weight by turning the screws. To find the force applied to lift a weight to a certain height, we must use the law of energy conservation. We know that if a system is isolated, then the total energy of the system remains constant. Which to say that, the energy of the system is conserved.
Formula used: $F\times 2\pi l=mga$

Here, let us consider that screw jack and the mass as an isolated system. Then from the law of conservation of energy, the energy of the system is conserved. We are spending some energy turning the screw. And we notice the mass being lifted to some height.
Thus clearly, the energy which is spent in turning the screw is used to lift the mass to a certain height. We also know that mechanical work is done to produce the required energy.
Then we get, $W_{s}=W_{m}$ where, $W_{s}$ is the work done in turning the screw and $W_{m}$ is the work done in lifting the mass. We know that $W=F\times d$ where, $d$ is the distance the body moves due to the applied force $F$.
Since we are turning the screw, $W_{s}=F\times 2\pi l$ where $2\pi l$ is the circumference of the handle of length covers. And $W_{m}=mga$ which is the force \[mg\] against the gravity to raise the mass by height $a$
Then $F\times 2\pi l=mga$
Given, $l=0.5m$ , $m=200kg$ and $a=5mm=5\times 10^{-3}m$
Then, $F=\dfrac{200\times 10\times 5\times 10^{-3}m}{ 2\times 3.14\times 0.5}=3.18N=3.2N$

Hence the answer is \[C.3.2N\]

Note:
The mechanical energy used in turning the screw is transferred to lift the object. Also greater the force, greater is the height to which the object is raised. Whereas the larger the handle, smaller is the force required to lift the object.