two masses m' each are hanging with help of a light string,which passes over mass less pulley. mass A starts leaking out at a rate of u kg/s with velocity v' with respect to mass A. find kinetic energy of block B as a function of time.
To find the kinetic energy of block B as a function of time, we need to consider the forces acting on both masses A and B.
Let's denote the masses of A and B as mA and mB, respectively. The rate at which mass A is leaking out is given as u kg/s, which means the mass of A is decreasing over time. Let's call the decreasing mass of A at any time t as mA(t).
The force acting on mass A is its weight, which is given by the formula F = m × g, where m represents the mass and g represents the acceleration due to gravity. Therefore, the force acting on A is mA(t) × g.
According to Newton's second law (F = m × a), the force acting on A is also equal to the mass of A multiplied by its acceleration. The acceleration of A is defined as the rate of change of its velocity (v') with respect to time. Hence, we can write:
mA(t) × g = mA(t) × d(v')/dt
Simplifying the equation, we get:
g = d(v')/dt
Now, let's consider the forces acting on mass B. The only force acting on B is the tension in the string. Since the string is light and passes over a massless pulley, the tension in the string is the same on both sides. Let's denote the tension in the string as T.
Using Newton's second law again, the equation for mass B becomes:
mB × g - T = mB × dvB/dt
Here, vB represents the velocity of mass B with respect to time. Notice that the acceleration term in this equation (dvB/dt) is not the same as the acceleration term for mass A since they have different motion characteristics.
Since the two sides of the string are moving with different velocities (v' for A and vB for B), we can relate them as:
vB = u + v'
Using this relation, we can rewrite our equation for B:
mB × g - T = mB × dvB/dt
mB × g - T = mB × d(u + v')/dt
mB × g - T = mB × du/dt + mB × dv'/dt
mB × g - T = mB × 0 + mB × dv'/dt
mB × g - T = mB × dv'/dt
Since we are finding the kinetic energy of B as a function of time, we need to integrate the above equation to eliminate the acceleration term (dv'/dt). By integrating both sides of the equation with respect to time (t), we get:
mB × g × t - ∫ T dt = ∫ mB dv'
The left-hand side of the equation represents the work done by the force against gravity (mB × g × t) minus the work done by the tension force (∫ T dt), whereas the right-hand side represents the change in kinetic energy of mass B.
Since T is constant and does not depend on time, the integral of T with respect to t simply becomes T × t.
Finally, the equation for the kinetic energy of mass B as a function of time (EkB(t)) becomes:
EkB(t) = mB × g × t - T × t + C
Here, C represents the constant of integration, which depends on the initial conditions of the system.
Therefore, the kinetic energy of mass B as a function of time is given by the equation:
EkB(t) = (mB × g - T) × t + C