Pulleys  (Page 5/5)

We are required to determine acceleration of block "B" to answer this question. By inspection, we observe that if block “A” moves “x” distance towards right, then the string length is distributed in two equal halves about hanging pulley. Clearly, if “A” moves by “x”, then “B” moves by x/2. As such, acceleration of “A” is twice that of “B”. As it is given that acceleration of “A” is 1 $m/s^2$ , the acceleration of “B” is 2 $m/s^2$ .

Hence, option (b) is correct.

Let the acceleration of block “A” is “a”. We know that the acceleration of block “A” is two times that of pulley. Hence,

$$a=1X2=2m/s^2$$

Let the tension in the string be “T”. Considering forces on block “A”, we have :

$$T=ma=0.5X2=1N$$

Considering forces on the mass-less pulley, we have :

$$\Rightarrow F=2T=2X1=2N$$

Hence, options (b) and (c) are correct.

Here, we need to determine relation of accelerations of blocks using constraint of string length.

$$x_A+x_A-x_0+x_B-x_0+x_B-x_0+x_C-x_0=L$$ $$\Rightarrow 2x_A+2x_B+x_C=L+3x_0=\text{constant}$$

Differentiating two times, we have :

$$\Rightarrow 2a_A+2a_B+a_C=0$$

Hence, option (a) is correct.

Let “T” be the tension in the string. Here, spring force is equal to tension in the string.

$$T=kx=200X0.1=20N$$

Let acceleration of block “A” is “a”. Considering forces on block “A”, we have :

$$\Rightarrow T=ma=10Xa$$

$$\Rightarrow a=\frac{T}{10}=\frac{20}{10}=2m/s^2$$

Considering forces on the mass-less pulley, we have :

$$\Rightarrow F=2T=2X20=40N$$

Now, acceleration of pulley is half of the acceleration of block. Hence, acceleration of pulley is 1 $m/s^2$ .

Hence, options (a) and (c) are correct.

Let the tensions in the strings connected to blocks “A” and “B” be “ $T_1$ ” and “ $T_2$ ” respectively. We see here that only force making block “A” to move on the table is tension in the string connected to it. Let acceleration of block “A” is “a”. From constraint relation, we know that the acceleration of “A” is twice the acceleration of “B”. Thus, acceleration of block “B” is “a/2”. Applying law of motion for block “A”, we have :

$$\Rightarrow T_1=m_{1}Xa=1Xa=a$$

Consideration of mass-less pulley,

$$\Rightarrow T_2=2T_1=2Xa=2a$$

Applying law of motion for block “A”, we have :

$$\Rightarrow m_{2}g-T_2=m_{2}X\frac{a}{2}$$ $$\Rightarrow 2X10-2a=2X\frac{a}{2}=a$$ $$\Rightarrow a=\frac{20}{3}m/s^2$$

Putting this value in the expressions of tensions, we have :

$$\Rightarrow T_1=a=\frac{20}{3}N$$ $$\Rightarrow T_2=2a=\frac{40}{3}N$$

Hence, options (b) is correct.

We consider blocks and string as one system. The net external force on the system is

$$F=nmg-mg=\left(n-1\right)mg$$

Total mass of the system, M, is :

$$M=nm+m=\left(n+1\right)m$$

The acceleration of the system i.e. the acceleration of either block is :

$$\Rightarrow a=\frac{\left(n-1\right)mg}{\left(n+1\right)m}=\frac{\left(n-1\right)g}{\left(n+1\right)}$$

According to question,

$$\Rightarrow a=\frac{\left(n-1\right)g}{\left(n+1\right)}=\frac{g}{3}$$ $$\Rightarrow 3n-3=n+1$$ $$\Rightarrow 2n=4$$ $$\Rightarrow n=2$$

Hence, option (d) is correct.

Let the acceleration of block “B” is “a”. Considering forces on block of mass “B”, we have :

$$F-T_1=Ma$$ $$\Rightarrow 2.5-T_1=1Xa=a$$

Considering free body diagram of mass-less pulley,

$$\Rightarrow T_1=2T_2$$

The pulley and the block are connected by inextensible string. As such, their accelerations are same. Further, we know by constraint analysis that acceleration of block is twice that of pulley. Hence, acceleration of block “B” is “2a”. Considering forces on block “A”, we have :

$$\Rightarrow T_2=mX2a=0.5X2a=a$$

Thus,

$$\Rightarrow T_1=2Xa=2a$$

Putting this value in the force equation of block “A”,

$$2.5-2a=0.5a$$ $$\Rightarrow a=1m/s^2$$

Clearly,

$$\Rightarrow T_1=2a=2X1=2N$$ $$\Rightarrow T_2=T_1/2=1N$$

Hence, option (a) is correct.