Physics 3 Marks Question

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.

1. Mass of man = 50kg.

g = 10m/s²

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

$\mu\text{R}+\mu\text{R = 40g}$

$\Rightarrow2\mu\text{R}=40\times10\Rightarrow\text{R}=\frac{40\times10}{2\times0.8}=250\text{N}$

2. The normal force is 250N.