Answer:
The value is [tex]KE = 259.6 \ J[/tex]
Explanation:
From the question we are told that
The weight of the horizontal solid disk is [tex]W = 805 \ N[/tex]
The radius of the horizontal solid disk is [tex]r = 1.58 \ m[/tex]
The force applied by the child is [tex]F = 49.5 \ N[/tex]
The time considered is [tex]t = 2.95 \ s[/tex]
Generally the mass of the horizontal solid disk is mathematically represented as
[tex]m_h = \frac{W}{ g}[/tex]
=> [tex]m_h = \frac{805}{ 9.8 }[/tex]
=> [tex]m_h = 82.14 \ N[/tex]
Generally the moment of inertia of the horizontal solid disk is mathematically represented as
[tex]I = \frac{1}{2} * m * r^ 2[/tex]
=> [tex]I = \frac{1}{2} * 82.14 * 1.58^ 2[/tex]
=> [tex]I = 102.5 \ kg \cdot m^2[/tex]
Generally the net torque experienced by the horizontal solid disk is mathematically represented as
[tex]T = I * \alpha = F * r[/tex]
=> [tex]\alpha = \frac{ F * r }{ I }[/tex]
=> [tex]\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }[/tex]
=> [tex]\alpha = 0.7628[/tex]
Gnerally from kinematic equation we have that
[tex]w = w_o + \alpha t[/tex]
Here [tex]w_o[/tex] is the initial angular velocity velocity of the horizontal solid disk which is [tex]w_o = 0\ rad/s[/tex]
So
[tex]w = 0 + 0.7628 * 2.95[/tex]
=> [tex]w = 2.2503 \ rad/s[/tex]
Generally the kinetic energy is mathematically represented as
[tex]KE = \frac{1}{2} * I * w^2[/tex]
=> [tex]KE = \frac{1}{2} * 102.53 * 2.2503 ^2[/tex]
=> [tex]KE = 259.6 \ J[/tex]
