A small box resting on one edge of the table | vedclass

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Question

(a)

For block, $v=0, t=2 s , s=1 m$

Now, by equation of motion, we have

$v=u+a t$

$0=u+a 	imes 2$

$\Rightarrow 2 a=-u \quad \dots(i)$

Vedclass · Accepted Answer

must be less than $0.05$

Vedclass · Answer

(a)

For block, $v=0, t=2 s , s=1 m$

Now, by equation of motion, we have

$v=u+a t$

$0=u+a 	imes 2$

$\Rightarrow 2 a=-u \quad \dots(i)$

$	ext { Now, } v^2-u^2=2 a s$

$	ext { Gives, }$

$\Rightarrow 0-u^2=-u 	imes 1$

$\Rightarrow u^2-u=0$

$\Rightarrow u(u-1)=0$

$	herefore u=0 	ext { or } u=1$

But $u=0$ is not a valid condition for given experiment, so

$u=1 \quad \dots(ii)$

As total initial energy is lost as work done against friction.

By conservation of energy, we have

$\frac{1}{2} m u^2=f_s$

$\Rightarrow \quad \frac{1}{2} m u^2=\mu_k m g_s$

$\Rightarrow \quad \frac{1}{2} \frac{u^2}{g s}=\mu_k$

$\Rightarrow \quad \mu_k=\frac{1}{2 	imes 10 	imes 1}=\frac{1}{20}$

$=0.05$

In above calculation, we does not consider the air resistance and other dissipative forces. So, $\mu_k$ must be slightly less than $0.05$ practically.

A small box resting on one edge of the table is struck in such a way that it slides upto the other edge, $1 \,m$ away after $2 \,s$. The coefficient of kinetic friction between the box and the table

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