3 Marks Question

Question 13 Marks

A person $(40\ kg)$ is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and the other wall B by his back. Assume that the friction coefficient between his body and the walls is $0.8$ and that limiting friction acts at all the contacts.

Show that the person pushes the two walls with equal force.
Find the normal force exerted by either wall on the person. Take $g = 10m/s^2.$

Answer

Mass of man $= 50\ kg.$

$g = 10m/s^2$
Frictional force developed between hands, legs back side with the wall the wt of man. So he remains in equilibrium.
He gives equal force on both the walls so gets equal reaction $R$ from both the walls. If he applies unequal forces $R$ should be different he can’t rest between the walls.
Frictional force $2\mu\text{R}$ balance his $wt.$
From the free body diagram
$\mu\text{R}+\mu\text{R = 40g}$
$\Rightarrow2\mu\text{R}=40\times10 \Rightarrow\text{R}=\frac{40\times10}{2\times0.8}=250\text{N}$
The normal force is $250N.$

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Question 23 Marks

A block is projected along a rough horizontal road with a speed of 10m/s. If the coefficient of kinetic friction is 0.10, how far will it travel before coming to rest?

Answer

Due to friction the body will decelerate
Let the deceleration be ‘a’
$\text{R}-\text{mg}=0\Rightarrow\text{R = mg} \ ...(1)$
$\text{ma}-\mu\text{R}=0\Rightarrow\text{ma}=\mu\text{R}=\mu\text{mg}$ (from(1))
$\Rightarrow\text{a}=\mu\text{g}=0.1\times10=1\text{m/s}^2$
Initial velocity u = 10m/s
Final velocity v = 0m/s
$\text{a}=-1\text{m/s}^2$
$\text{S}=\frac{\text{v}^2-\text{u}^2}{2\text{a}}=\frac{0-10^2}{2(-1)}=\frac{100}{2}=50\text{m}$
It will travel 50m before coming to rest.

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Physics 3 Marks Question

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