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1.

A stone is released from the top of a tower of height 19.6m. Calculate its final velocity just before touching the ground.

Answer» Correct Answer - 19.6 m/s
According to the equation of motion under gravity :
` v^(2)-u^(2)=2gs`
where,
u- initial velocity of the stone =0
v= Final velocity of the stone
s= Height of the stone `=19.6 m`
g= Acceleration due to gravity `=9.8 ms^_(-2)`
` therefore v^(2)-0^(2)=2xx9.8 xx19.6xx19.6`
`v^(2)=2xx9.8xx19.6 (19.6)^(2)`
` v=19.6 ms^(-1)`
Hence , the velcoity of the stone just before toching the ground is 19.6 `ms^(-1)`
Discussion
2.

A block of wood is kept on a table top The mass of the wooden block is 5 kg and its dimensions are `40cmxx20cmxx10cm`. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table with its sides of dimension (a)`20cmxx10cm`(b) `40cmxx20cm`. Given`g=9.8(m)/(s^2)`.

Answer» the mass of the wooden block =5 kg
the dimensions
`=40 cm xx 20 cm xx 10 cm `
Here , the weight of the wooden block applies a thrust on the table top .
that is
Thrust `= F = mxxg`
`=5 kg xx9.8 m s^(-2)`
=49 N
Area of side = length`xx` breadth
`=20 cm xx 10 cm `
`=200 cm ^(2) = 0.02 m^(2)`
From Eq .(10.20)
Pressure `=(49)/( 0.02 m^(2)) `
`= 2450 N m^(-2)`
when th block lies on its side of dimensions `40 cm xx20 cm ` it exerts the same throust
Aera = length `xx` breadth
`=40 cm xx20 cm`
`=800 cm ^(2) = 0.08 m^(2)`
From Eq .(10.20) ,
pressure `=(49) /(0.08 m^(2))`
=612.5 ` N m^(-2)`
the pressure ecerted by the side` 20 cm xx 10 cm ` is ` 2450 N m^(2) ` and by th side `40 cm xx 20cm ` is 612.5 `N m ^(-2)`
Discussion