Two blocks with masses m1=3 kg and m2=5 kg are connected by a light string that slides over a frictionless pulley as shown in figure. Initially, m2, is held 5 m off the floor while m1 is on the floor. The system is then released. At what speed (in m / s ) does m2 hit the floor?

Question

Two blocks with masses m1=3 kg and m2=5 kg are connected by a light string that slides over a frictionless pulley as shown in figure. Initially, m2, is held 5 m off the floor while m1 is on the floor. The system is then released. At what speed (in m / s ) does m2 hit the floor? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The initial and final configurations are shown in the figure. It is convenient to set

U = 0

at the floor. Initially, only

m_{2}

has potential energy. As it falls, it loses potential energy and gains. kinetic energy. At the same time,

m_{1}

gains potential energy and kinetic energy. Just before

m_{2}

lands, it has only kinetic energy. Let

v

the final speed of each mass. Then, using the law of conservation of mechanical energy.
Final (Just before

m_{2}

strikes the floor)

K_{f} + U_{f} = K_{i} + U_{i}

\frac{1}{2} (m_{1} + m_{2}) v^{2} + m_{1} g h = 0 + m_{2} g h

v^{2} = \frac{2 ( m _{2} - m _{1} ) g h}{m _{1} + m _{2}}

Putting

m_{1} = 3 k g; m_{2} = 5 k g

;

h = 5 m

and

g = 10 m / s^{2}

we get

v^{2} = \frac{2 ( 5 - 3 ) ( 10 ) ( 5 )}{5 + 3}

or

v = 5 m / s

Q. Two blocks with masses m1​=3kg and m2​=5kg are connected by a light string that slides over a frictionless pulley as shown in figure. Initially, m2​, is held 5m off the floor while m1​ is on the floor. The system is then released. At what speed (in m/s ) does m2​ hit the floor?