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Step-by-Step Solution
Step 1: Write down the known information
• Mass of the body, m = 2 \text{ kg} .
• The source provides constant power P .
• The body moves from rest ( v = 0 at t=0 ).
• Displacement after 4 seconds is given as s = \frac{1}{3}\,\alpha^{2}\,\sqrt{P}\,\text{m} .
Step 2: Relationship between power and velocity
Under constant power P , the power is related to force F and velocity v by
P = F \, v . Since F = m \frac{dv}{dt} , we can write:
P = m \, v \, \frac{dv}{dt}.
Step 3: Express velocity as a function of time
Rewrite the above differential equation:
m \, v \, \frac{dv}{dt} = P.
Rearrange and integrate from v=0 to v and t=0 to t :
\int_{0}^{v} m \, v \, dv = \int_{0}^{t} P \, dt.
This becomes:
\frac{m \, v^{2}}{2} \Bigg|_{0}^{v} = P\,t \Bigg|_{0}^{t}.
Hence,
\frac{m\,v^{2}}{2} = P\,t \quad \Longrightarrow \quad v = \sqrt{\frac{2P}{m}}\;\sqrt{t}.
Step 4: Find the displacement as a function of time
Displacement s is obtained by integrating velocity over time:
s = \int_{0}^{t} v \, dt = \int_{0}^{t} \sqrt{\frac{2P}{m}}\,\sqrt{t}\, dt.
Factor out constants:
s = \sqrt{\frac{2P}{m}} \int_{0}^{t} t^{\frac{1}{2}}\, dt = \sqrt{\frac{2P}{m}} \;\times\; \frac{2}{3}\, t^{\frac{3}{2}}.
Therefore,
s = \frac{2}{3} \,\sqrt{\frac{2P}{m}}\, t^{\frac{3}{2}}.
Step 5: Substitute m=2\,\text{kg} and t=4\,\text{s}
Substitute m=2 and t=4 :
\sqrt{\frac{2P}{m}} = \sqrt{\frac{2P}{2}} = \sqrt{P},
and
t^{\frac{3}{2}} = 4^{\frac{3}{2}} = 4^{1}\times 4^{\frac{1}{2}} = 4 \times 2 = 8.
Hence,
s = \frac{2}{3} \,\sqrt{P}\,\times 8 = \frac{16}{3} \sqrt{P}.
Step 6: Compare with the given displacement form
The problem statement says
s = \frac{1}{3} \,\alpha^{2}\,\sqrt{P}.
We have found
s = \frac{16}{3} \sqrt{P}.
Equating:
\frac{16}{3} \sqrt{P} = \frac{1}{3}\,\alpha^{2}\,\sqrt{P}.
Cancel \frac{1}{3}\sqrt{P} on both sides:
16 = \alpha^{2} \quad \Longrightarrow \quad \alpha = 4.
Final Answer
\alpha = 4
