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 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:
Show 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.
Updated On: Jun 24, 2026
- \(24\,\text{J}\)
- \(31\,\text{J}\)
- \(30\,\text{J}\)
- \(25\,\text{J}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use the work-energy theorem.
According to the work-energy theorem, \[ W=\Delta K \] where \[ W=\text{work done} \] and \[ \Delta K=K_f-K_i \]
Step 2: Find the work done by the force.
Given, \[ \vec{F}=4\hat{i}-15\hat{j} \] and displacement \[ \vec{s}=6\hat{i} \] Work done is the dot product: \[ W=\vec{F}\cdot \vec{s} \] \[ =(4\hat{i}-15\hat{j})\cdot (6\hat{i}) \] \[ =24 \] Thus, \[ W=24\,\text{J} \]
Step 3: Find the final kinetic energy.
Initial kinetic energy: \[ K_i=7\,\text{J} \] Using \[ W=K_f-K_i, \] we get \[ 24=K_f-7 \] \[ K_f=31\,\text{J} \]
Step 4: Final conclusion.
Hence, the kinetic energy at the end of the displacement is \[ \boxed{31\,\text{J}} \]
According to the work-energy theorem, \[ W=\Delta K \] where \[ W=\text{work done} \] and \[ \Delta K=K_f-K_i \]
Step 2: Find the work done by the force.
Given, \[ \vec{F}=4\hat{i}-15\hat{j} \] and displacement \[ \vec{s}=6\hat{i} \] Work done is the dot product: \[ W=\vec{F}\cdot \vec{s} \] \[ =(4\hat{i}-15\hat{j})\cdot (6\hat{i}) \] \[ =24 \] Thus, \[ W=24\,\text{J} \]
Step 3: Find the final kinetic energy.
Initial kinetic energy: \[ K_i=7\,\text{J} \] Using \[ W=K_f-K_i, \] we get \[ 24=K_f-7 \] \[ K_f=31\,\text{J} \]
Step 4: Final conclusion.
Hence, the kinetic energy at the end of the displacement is \[ \boxed{31\,\text{J}} \]
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