Question:
A force of \(5\,\text{N}\) acts on a body initially at rest. If the instantaneous power due to the force at the end of the third second \((t=3\,\text{s})\) is \(5\,\text{W}\), then the mass of the body is
A force of \(5\,\text{N}\) acts on a body initially at rest. If the instantaneous power due to the force at the end of the third second \((t=3\,\text{s})\) is \(5\,\text{W}\), then the mass of the body is
Show Hint
Instantaneous power delivered by a force is
\[
P=\vec{F}\cdot \vec{v}
\]
For motion along the same direction,
\[
P=Fv
\]
directly.
Updated On: Jun 26, 2026
- \(25\,\text{kg}\)
- \(12.5\,\text{kg}\)
- \(15\,\text{kg}\)
- \(7.5\,\text{kg}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Use the relation between force and acceleration.
Given force, \[ F=5\,\text{N} \] Using Newton's second law, \[ F=ma \] So, \[ a=\frac{F}{m} \] \[ a=\frac{5}{m} \]
Step 2: Find velocity at the end of \(3\,\text{s}\).
The body starts from rest, so \[ u=0 \] Using \[ v=u+at \] At \[ t=3\,\text{s}, \] we get \[ v=0+\frac{5}{m}(3) \] \[ v=\frac{15}{m} \]
Step 3: Use the formula for instantaneous power.
Instantaneous power is \[ P=Fv \] Given, \[ P=5\,\text{W} \] Thus, \[ 5=5\left(\frac{15}{m}\right) \] \[ 5=\frac{75}{m} \] \[ m=\frac{75}{5} \] \[ m=15\,\text{kg} \]
Step 4: Final conclusion.
Hence, the mass of the body is \[ \boxed{15\,\text{kg}} \]
Given force, \[ F=5\,\text{N} \] Using Newton's second law, \[ F=ma \] So, \[ a=\frac{F}{m} \] \[ a=\frac{5}{m} \]
Step 2: Find velocity at the end of \(3\,\text{s}\).
The body starts from rest, so \[ u=0 \] Using \[ v=u+at \] At \[ t=3\,\text{s}, \] we get \[ v=0+\frac{5}{m}(3) \] \[ v=\frac{15}{m} \]
Step 3: Use the formula for instantaneous power.
Instantaneous power is \[ P=Fv \] Given, \[ P=5\,\text{W} \] Thus, \[ 5=5\left(\frac{15}{m}\right) \] \[ 5=\frac{75}{m} \] \[ m=\frac{75}{5} \] \[ m=15\,\text{kg} \]
Step 4: Final conclusion.
Hence, the mass of the body is \[ \boxed{15\,\text{kg}} \]
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