Using the work-energy theorem, the total work done by all forces is equal to the change in kinetic energy (\(\Delta KE\)) of the system. Here:
\[w_g + w_{Fr} + w_s = \Delta KE.\]
Substituting the given values:
\[5 \times 10 \times 5 - 0.5 \times 5 \times 10 \times x - \frac{1}{2} Kx^2 = 0 - 0.\]
Simplifying:
\[250 - 25x - 50x^2 = 0.\]
Rewriting:
\[2x^2 + x - 10 = 0.\]
Solving this quadratic equation gives:
\[x = 2.\]
Thus, the maximum compression in the spring is \(x = 2\) m.
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).