Question

How much work is done to lift a $20kg$ sack of potatoes vertically $6.5m$?

Answer 1

Hint: The work done on a body is the product of force vector and displacement vector. It is the scalar product of two vectors and hence is a scalar quantity. When the sack of potatoes is lifted, force of gravity acts on it and the height is its displacement. Substituting values in the formula for work done, work can be calculated.

Complete answer:
The product of force acting on an object and the displacement travelled by it is known as work. Its SI unit is joules ($J$). It is a scalar quantity.
$W=\vec{F}\cdot \vec{s}$
Here, $W$ is the work done
$\vec{F}$ is the force vector
$\vec{s}$ is the displacement vector

From the above equation,
$W=Fs\cos \theta $ …………………. (1)
Here, $\theta $ is the angle between force and displacement vector

When a sack of potatoes is lifted vertically upwards, it is lifted against the force of gravity acting on it. The force of gravity on $20kg$ sack will be-
$F=mg$
Here, $F$ is the force of gravity
$m$ is the mass of a body
$g$ is the acceleration due to gravity

Substituting given values for the sack of potatoes in the above eq, we get,
$\begin{align}
  & F=20\times 10 \\
 & \Rightarrow F=200N \\
\end{align}$

Therefore, the force acting on the sack of potatoes is $200N$. It is lifted to a height of $6.5m$, therefore, its displacement is $6.5m$. Angle between force of gravity and displacement is ${{180}^{o}}$. Substituting given values in eq (1), we get,
$\begin{align}
  & W=200\times 6.5\cos {{180}^{o}} \\
 & \Rightarrow W=-1300J \\
\end{align}$

Therefore, the work done by the force of gravity on the sack of potatoes is $-1300J$.

Note:
The work done is negative because it is done against the force of gravity. Work done is stored in a body as energy. Similarly, this work done is stored in the body as potential energy. The higher the energy of a body, the more unstable it is. Also work done equal to the potential energy is to be done on the sack of potatoes by an external force to lift it against the force of gravity.