A uniform rectangular block of wood 20 cm x 7.5 cm x 7.5 cm and of mass 1000g stands on a horizontal surface with its longest edge vertical. Find (i) the pressure exerted by the block on the surface (ii) the density of the wood. Ans. (1778 Nm⁻², 889 kgm⁻³)


Given:  

Volume of the rectangular wooden block (from the measurement) = V =  20 cm ｘ 7.5 cm ｘ 7.5 cm = 1,125 cm³ = 1.125 ｘ1o³ cm³ = 1.225 ｘ1o³ ｘ1o⁻⁶ m³ = 1.125 ｘ1o⁻³ m³ ( 1 cm³ = 1o⁻⁶ m³)

Area of the rectangular wooden block (from the measurement) = A = 7.5 cm ｘ 7.5 cm = 56.25 cm² = 5.625 ｘ1o cm² = 5.625 ｘ1o ｘ1o⁻⁴ m² = 5.625 ｘ1o⁻³ m² ( 1 cm² = 1o⁻⁴ m)

Mass of the block  = m =  1000 g = 1 kg
Gravitational Acceleration  = g = 10 m s⁻²

To Find:

(i) the pressure exerted by the block on the surface = P = ?
(ii) density of the wood = 𝜌  = ?

Solution:   

(i) the pressure exerted by the block on the surface = P = ?

As Pressure P is defined as force per unit area. So, 

P = `\frac {F}{A}`

Here Force F = weight of the block = mg so, 

P = `\frac {mg}{A}`

 by putting values

P = `\frac {1 kg ｘ 10 ms⁻²}{5.625 ｘ10⁻³ m²}`

P = 0.1777ｘ10⁴ kg m s⁻²m⁻²  (kg m s⁻² = N)

by putting (kg m s⁻² = N)

P = 0.1777ｘ10⁴ N m⁻²

or

P = 17777 N m⁻²

thus, the pressure extorted by the block on the surface is 17777 N m⁻²



(ii) density of the wood = 𝜌  = ?

As Density is defined as mass per unit volume. So, 

Density = `\frac {Mass}{Volume}`

𝜌 = `\frac {m}{V}`

 by putting values

𝜌 = `\frac {1 Kg}{1.125 ｘ1o⁻³ m³}`

𝜌 = 0.889 ｘ1o³ Kgm⁻³

or

𝜌 = 889 Kgm⁻³

Thus, the density (𝜌) of the wood of the block is 889 Kgm⁻³

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