A 5 kg wooden block is kept on a table top. If its
dimensions are 40 cm x 30 cm x 20 cm, find the
pressure exerted by the wooden block on the table
top, if it is made to lie on the table top with its
sides of dimensions
(i) 40 cm x 20 cm
(ii) 40 cm x 30 cm
Answers
Answer:
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sharinkhan
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mass of the wooden block = m= 5 kgdimensions = 40 cm × 20 cm × 10 cm
the weight of the wooden block applies a thrust on the table top.That is,Thrust = F = mg= 5 kg × 9.8 m/s²= 49 N
Area of a side = length × breadth
= 20 cm × 10 cm
= 200 cm²
= 0.02 m²
Pressure = thrust/ area
Pressure = 49 /0.02= 2450 N/m²
The same thrust in implied when it lies on the given dimension.
Area= length × breadth= 40 × 20= 800= 0.08 m²
Pressure = 49/0.08= 612.5
Explanation:
1. We know, Pressure = Force/ area
So, Pressure = 50000/40*20
= 50000/800
= 62.5 Pa
2. We know, Pressure = Force/ area
So, Pressure = 50000/40*30
= 50000/1200
= 41.67 Pa
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