A force \(F =\left(5+3 y^2\right)\) acts on a particle in the \(y\)-direction, where \(F\) is in newton and \(y\) is in meter The work done by the force during a displacement from \(y=2 m\) to \(y=5 m\) is___ \(J\).
Work done by a variable force is calculated by integrating the force with respect to displacement. Ensure the force and displacement are in the same direction.
Step 1: Recall the Formula for Work Done
The work done (\(W\)) by a variable force \(F(y)\) over a displacement from \(y_1\) to \(y_2\) is given by the integral:
\[ W = \int_{y_1}^{y_2} F(y) \, dy \]
Step 2: Substitute the Given Force and Limits
In this case, \(F(y) = 5 + 3y^2\), \(y_1 = 2 \, \text{m}\), and \(y_2 = 5 \, \text{m}\). So,
\[ W = \int_{2}^{5} (5 + 3y^2) \, dy \]
Step 3: Evaluate the Integral
\[ W = \left[ 5y + \frac{3y^3}{3} \right]_2^5 = \left[ 5y + y^3 \right]_2^5 \] \[ W = (5(5) + 5^3) - (5(2) + 2^3) \] \[ W = (25 + 125) - (10 + 8) \] \[ W = 150 - 18 = 132 \, \text{J} \]
Conclusion: The work done by the force is \(132 \, \text{J}\).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
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