Question

How much does it take to lift a 35kg weight $\dfrac{1}{2}$ meter?

Answer 1

Hint: Work is the energy transferred to or from an object by the application of force along with a displacement. In its simplest form, it is often represented as the product of force and displacement. A force is said to do positive work if (when applied) it has a component in the direction of the displacement of the point of application. A force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force. The gravitational force on a body, Newton’s second law, is defined as the product of the gravitational acceleration and the mass of the body.

Complete solution:
To find the work done to lift the 35kg weight to the half meter first find out the gravitational force on the 35kg. Express the relation between gravitational force and the weight of the body.
$\therefore {F_g} = m.g$ , where $g$ is the acceleration due to gravity.
Substitute the values 35kg for $m$ , $9.8m/{s^2}$ for $g$
$\therefore {F_g} = - 35 \times 9.8$
$ \Rightarrow {F_g} = - 343N$
Express the formula for the work to lift the weight from the earth's surface.
$\therefore W = {F_g}d$ , where $d$ is the displacement given in the question as $d = 0.5m$.
$\therefore W = - 343 \times 0.5$
$ \Rightarrow W = - 171.5J$
The unit of work is Joule.

Note: Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to $mgh$ on it, thereby increasing its kinetic energy by that same amount. Every object attracts another object in the universe therefore when an object is in the gravitational field of another object then it stores some potential energy in the form of the force exerted by the other object. To lift the object, we have to work against the force of attraction between them. Therefore, the sign of force applied to lift an object is negative, and eventually, the above result becomes negative.