Question

An object is displaced from position vector ${r_1} = \left( {4i + 6j} \right)\;{\text{m}}$ under the action of a force $F = \left( {3{x^2}i + 2yj} \right)\;{\text{N}}$. Find the work done by this force.

Answer 1

The work done by the force is the same as the work to displace the object by force and the displacement of the object. Use the dot product of the force vector and position vector of the object to find the force's work. The work done by a force is the scalar quantity.

Complete step by step answer:
Given, The position vector of the object is ${r_1} = \left( {4i + 6j} \right)\;{\text{m}}$, the force on the object is $F = \left( {3{x^2}i + 2yj} \right)\;{\text{N}}$.
The equation to calculate the work done by the force is,
$W = F \cdot {r_1}......\left( 1 \right)$
Substitute ${r_1} = \left( {4i + 6j} \right)\;{\text{m}}$ and $F = \left( {3{x^2}i + 2yj} \right)\;{\text{N}}$ in the equation (1).
$\Rightarrow W = \left[ {\left( {3{x^2}i + 2yj} \right)\;{\text{N}}} \right]\left[ {\left( {4i + 6j} \right)\;{\text{m}}} \right]$
$\Rightarrow W = \left( {12{x^2} + 12y} \right)\;{\text{N}} \cdot {\text{m}}$

Therefore, the work done by the force is $\left( {12{x^2} + 12y} \right)\;{\text{N}} \cdot {\text{m}}$.

Additional Information:
The product of the same unit vector components becomes one, and the product of different unit vectors becomes zero. The dot product of the two vector quantities produces a scalar quantity. The force and position of the object are vector quantity, so the work is the scalar quantity. The work done depends on the initial and final position of the object and type of force. If the force is conservative, then the work becomes path independent. If the work is non-conservative, then the work becomes path-dependent.

Note:
Be careful in calculating the dot products of force and position vector to find the work. The component of the force only along the displacement is considered to calculate the work done by the force.