Question

A cubical block of wood of length 5 cm is kept on a table top. The density of the block is 800 kg/m2 and g 9.8 ms. Find the pressure exerted by the wooden block on the table top.​

Answer 1

volume of cubical block, V = (side length)³

= (5cm)³ = 125cm³

we know, 1 cm³ = 10^-6 m³

so, volume of cubical block, V = 1.25 × 10^-4 m³

given, density of block , d = 800 kg/m³

so, mass of block , m = Vd

= 800 × 1.25 × 10^-4

= 0.1 kg

now, pressure exerted by the wooden block on the table top , P = W/A

where A is area of each face of cubical block. i.e., A = (5 × 10^-2 m)² = 2.5 × 10^-3 m²

and W is weight of wooden block.i.e., W = mg = 0.1 × 9.8 = 0.98 N

hence, P = 0.98/(25 × 10^-3)

= 0.98 × 10³/25

= 980/25

= 39.2 N/m²