Question

A force \[ \vec{F}=4\hat{i}-15\hat{j}\ \text{N} \] acts on a body resulting in a displacement of \[ 6\hat{i}. \] If the body had a kinetic energy of \(7\,\text{J}\) at the beginning of the displacement, then the kinetic energy at the end of the displacement is:

• A. \(24\,\text{J}\)
• B. \(31\,\text{J}\)
• C. \(30\,\text{J}\)
• D. \(25\,\text{J}\)

Hint:

Work done by a force is calculated using the dot product: \[ W=\vec{F}\cdot \vec{s}. \] Only the component of force along the displacement contributes to work.

Answer 1

The Correct Option is B

Step 1: Recall the work-energy theorem.
The work done by all forces on an object equals the change in its kinetic energy:
\[ W_\text{net} = K_f - K_i \]

Step 2: Identify which force components do work.
The displacement is $\vec{s} = 6\hat{i}$, which is purely in the $x$-direction.
Work is done only by the component of force parallel to the displacement.
So only $F_x = 4\,\text{N}$ does work; the $y$-component ($-15\hat{j}$) does no work because it is perpendicular to the displacement.

Step 3: Compute the work done by the x-component of force.
\[ W = F_x \times s_x = 4 \times 6 = 24\,\text{J} \]

Step 4: Verify using dot product.
\[ W = \vec{F} \cdot \vec{s} = (4\hat{i} - 15\hat{j}) \cdot (6\hat{i}) = 24 + 0 = 24\,\text{J} \]

Step 5: Apply the work-energy theorem.
Initial kinetic energy $K_i = 7\,\text{J}$.
\[ W = K_f - K_i \implies K_f = K_i + W = 7 + 24 = 31\,\text{J} \]

Step 6: State the final answer.
\[ \boxed{31\,\text{J}} \]