First slide

Second law

Question

Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed $υ$ . What is the net force on the block of mass 2m? (g is the acceleration due to gravity)

Moderate

A

3 mg
B

6 mg
C

Zero
D

2 mg

Solution

Let $T_{1}$ be tension in string connecting m and 2m and $T_{2}$ be tension in string connecting 2m and 3m.
Let a be common acceleration of the system.
$∴ a = \frac{F - (m + 2 m + 3 m) g}{m + 2 m + 3 m} = \frac{F - 6 m}{6 m}$
As the system moves with constant speed, therefore, a = 0
$∴ F - 6 m g = 0 o r F = 6 m g$
The free body diagram of block m is as shown in the figure.
The equation of motion of block of mass m is
$F - T_{1} - m g = 0 6 m g - T_{1} - m g = 0 T_{1} = 5 m g - - - (i)$
The free body diagram of block of mass 2m is as shown in the figure.
The equation of motion of block of mass 2m is
$T_{1} - T_{2} - 2 m g = 0 5 m g - T_{2} - 2 m g = 0 (U \sin g (i)) T_{2} = 3 m g$
The free body diagram of block of mass
3m is as shown in the figure.
The equation of motion of block of mass 3m is
$T_{2} - 3 m g = 0 T_{2} = 3 m g$
Net force on the block of mass 2m is
$F_{n e t} = T_{1} - T_{2} - 2 m g = 5 m g - 3 m g - 2 m g = 0$
Alternate solution
As all blocks are moving with constant speed, therefore, acceleration is zero. So net force on each block is zero.

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