Question

How much work is it to lift a \[20\;{\rm{kg}}\] sack of potatoes vertically \[6.5\;{\rm{m}}\]?

Answer 1

Hint: The above problem can be resolved using the mathematical relation for the work done to lift the object upto a specific height. This work done is stored in the form of the potential energy of the object. Hence the equation of the work done involves the variables like the mass of an object, height upto which the object is being lifted and the constant value of the gravitational acceleration, which upon substituting the values one will obtain the desired result.

Complete step by step answer:
Given:
The mass of the sack is, \[m = 20\;{\rm{kg}}\].
The vertical distance upto which the sack is lifted is, \[h = 6.5\;{\rm{m}}\].
The work done to lift the sack is given as,
\[W = m \times g \times h\]
Here, g is the gravitational acceleration and its value is \[9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\].
Solve by substituting the values in the above equation as,
\[\begin{array}{l}
W = m \times g \times h\\
W = 20\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \times 6.5\;{\rm{m}}\\
W = 1274\;{\rm{J}}
\end{array}\]
Therefore, the required work to lift the sack is \[1274\;{\rm{J}}\].

Note:To resolve the given numerical, one must be clear about the basic condition that when an object is being raised to some particular height, then there occurs the entire work to be stored in the form of the potential energy of the object. The potential energy is that form of energy, which is incorporated as the energy due to the height. Moreover, the fundamental of the type of work is to be analysed, as if the displacement happening in the direction of applied force, then the work is known as positive work and if the displacement is in the reverse direction, then work obtained is called negative work done.