Question

Block A is weighing 20 kg is placed on a smooth surface. Weight B of 2 kg is mounted on the block. The coefficient of friction between the block and the weight is 0,25. Calculate the acceleration of the frictional force between the block and the weight when a horizontal force of 2 N is applied to the weight as shown in Fig. What will these quantities be if the horizontal force is 20N? (g= 10 $m{s}^{-2}$)

• A. $\frac{1}{21}m{s}^{-2};\frac{20}{11}Nm{s}^{-2};0.25m{s}^{-2},5N$
• B. $\frac{1}{11}m{s}^{-2};\frac{20}{11}Nm{s}^{-2};0.25m{s}^{-2},5N$
• C. $\frac{1}{11}m{s}^{-4};\frac{20}{11}Nm{s}^{-2};0.25m{s}^{-2},6N$
• D. $\frac{1}{11}m{s}^{-5};\frac{20}{11}Nm{s}^{-2};0.25m{s}^{-2},7N$

Answer 1

The correct option is B $\frac{1}{11}m{s}^{-2};\frac{20}{11}Nm{s}^{-2};0.25m{s}^{-2},5N$

So long as the applied force does not produce the limiting frictional force, the two bodies will have on relative motion than is the two will behave in relative motion. Here,

$\text{limiting friction}=0.5\times 2\times 10=5N$

Let $F$ be the force that produces limiting friction between the two. Then,

$F-5=2a$

$5=20a$ where, $a=$ common acceleration.

hence,

$F=5.5N$

Here, the applied force is only 2N. Hence,the two bodies will behave as one and will have a common acceleration given by $2=(20+2)a\text{or,}a=\frac{1}{11}m{s}^{-2}$

$F=ma=20\times \frac{1}{11}=\frac{20}{11}N$

$\therefore F-f=2a_1$

$\Rightarrow 20-5=2a_1$

$a_1=7.5m{s}^{-2}$, $F=5N$

Also; $5=20a_2$

$a_2=0.25m{s}^{-2}$