Question

② Derive expression for displacement and time taken
for a body to come to rest on a rough horizontal
surface?​

Answer 1

Derivation:

• Let's assume that initial velocity of body is u, coefficient of friction is $\mu$, and mass of body is m.

Now, let time taken to stop be t:

$\sf \: v = u + at$

$\sf \implies\: 0 = u +( - \mu g)t$

$\sf \implies\: 0 = u - \mu gt$

$\sf \implies\: t = \dfrac{u}{ \mu g}$

Now, displacement be d :

$\sf \: {v}^{2} = {u}^{2} + 2ad$

$\sf \implies\: {0}^{2} = {u}^{2} + 2( - \mu g)d$

$\sf \implies\: 2 \mu gd = {u}^{2}$

$\sf \implies\: d = \dfrac{{u}^{2} }{2 \mu g}$

Hope It Helps.

Answer 2

Explanation:

$Derivation:</p><p></p><p>Let's assume that initial velocity of body is u, coefficient of friction is \muμ , and mass of body is m.</p><p></p><p>Now, let time taken to stop be t:</p><p></p><p>\sf \: v = u + atv=u+at</p><p></p><p>\sf \implies\: 0 = u +( - \mu g)t⟹0=u+(−μg)t</p><p></p><p>\sf \implies\: 0 = u - \mu gt⟹0=u−μgt</p><p></p><p>\sf \implies\: t = \dfrac{u}{ \mu g}⟹t=μgu</p><p></p><p>Now, displacement be d :</p><p></p><p>\sf \: {v}^{2} = {u}^{2} + 2adv2=u2+2ad</p><p></p><p>\sf \implies\: {0}^{2} = {u}^{2} + 2( - \mu g)d⟹02=u2+2(−μg)d</p><p></p><p>\sf \implies\: 2 \mu gd = {u}^{2}⟹2μgd=u2</p><p></p><p>\sf \implies\: d = \dfrac{{u}^{2} }{2 \mu g}⟹d=2μgu2</p><p></p><p>Hope It Helps.</p><p></p><p>$