Question

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Answer 1

Answer:

$a_t=\frac{F(3m1+2m_2)}{m_1(m_1+m_2)}\\\;\\{KE}=\frac{F^2t^2(3m_1+2m_2)}{2m_1(m_1+m_2)}$

Explanation: In this question we will calculate all the data and terms around horizontal displacement and rotations because by applying the force F, mass m₁ will start rotating while mass (m₁ + m₂) will(may) move in horizontal direction by this force.

Let the acceleration of the body in horizontal direction i.e. for linear displacement be a and Also let acceleration  for rotation of of m₁ be α i.e. angular acceleration of the mass m₁

F will be responsible for movement of mass (m₁ + m₂) that will give them acceleration of 'a'

Thus,

F = (m₁ + m₂)a

a = F/(m₁ + m₂)

We know that inertia of a solid cylinder with radius r is given by,

I = m₁r²/2

Now,

torque  = I × α

F × r = I × α

F × r  = m₁r²/2 × α

2F/m₁ = αr

αr = 2F/m₁

We know that total acceleration is given by

$a_t=a+\alpha r\\\;\\a_t=\frac{F}{m_1+m_2}+\frac{2F}{m_1}\\\;\\a_t=\frac{F(m_1+2m_1+2m_2)}{m_1(m_1+m_2)}\\\;\\a_t=\frac{F(3m_1+2m_2)}{m_1(m_1+m_2)}\\\;\\\textbf{This will be total acceleration of the system}$

b) Since body was initially in the rest i.e. u = 0

Let the displacement of the body along x-axis be x.

Now from the second equation of motion,

$x=u+\frac{1}{2}a_tt^2\\\;\\x=\frac{1}{2}[\frac{F(3m_1+2_2)}{m_1(m_1+m_2)}]t^2\\\;\\x=\frac{Ft^2(3m_1+2m_2)}{2m_1(m_1+m_2)}$

Since Kinetic energy of the system will be equation to  total work done one the body.

Therefore,

$\text{KE}=\text{Force}\times\text{Displacement}\\\;\\\text{KE}=F\times x\\\;\\\text{KE}=F\times\frac{Ft^2(3m_1+2m_2)}{2m_1(m_1+m_2)}\\\;\\\text{KE}=\frac{F^2t^2(3m_1+2m_2)}{2m_1(m_1+m_2)}$