Question

If 100 N force is applied to the 10 kg block as shown in diagram. What is the acceleration produced for slab
and block?
plz no irrelevant answers.
I want to know for the block.
​

Answer 1

$a_{slab}\:=\:0.98ms^{-2}$

$a_{block}\:=\:6.08ms^{-2}$

Given:-

• A block of mass 10 kg kept over a slab of mass 40 kg.

• A force of 100 N is applied to the block.

• $\mu_S=0.6$,
• $\mu_K=0.4$

To find:-

Acceleration produced for slab and block

Solution:-

From Fig. 1,

There is an unbalanced force, F = 100 N in the horizontal direction.

$\therefore$Movement occurs.

Let's assume that both the block and the slab move together with a common acceleration.

$a_c=\dfrac{100}{10+40}=2m/s^2$

Required frictional force, f = 40 × 2 = 80 N

Now limiting friction or the maximum static friction available,

$f_L=\mu N_1\: = 0.6\times10\times10=20\:N$

Clearly, f exceeds $f_L$

$\implies$The block and the slab move separately.

Acceleration of the slab,

F = ma or

$a = \dfrac{F}{m}$

$=\dfrac{\mu_K\times9.8\times10}{40}$

$a_{slab}=0.98\:ms^{-2}$

Acceleration of the block,

Net force acting in the direction of the block;

= F - $\mu$mg

= 100 - 0.4×10×9.8

= 100 - 39.2

= 60.8 N

Now, $a_{block}=\dfrac{F}{m}$

=$\dfrac{60.8}{10}$

= 6.08 m/s²