H= 4m, l=15m,a=800cm2,d=2.5×10 at rest 3 kg/m3,g=10m/s2 . find work done?
Answers
Answer:
Explanation:
Let P
1
,P
2
and P
3
be the pressures exerted by the brick while resting on different faces.
The dimensions of the given brick are 20cm×10cm×5cm
Case (i) : When the block is resting on 20cm×10cmface.
Thrust acting= Weight of the brick
T=500gwt
Area of constant (A)=20cm×10cm
Pressure exerted (P
1
)
=
Area
Thrust
=
20×10
500
∴P
1
=2.5gwtcm
−2
Case (ii) : When the block is resting on 20cm×5cmface
Thrust= Weight of the brick
=500gwt
Area of constant (A)=20cm×5cm
Pressure exerted
(P
2
)=
Area
Thrust
=
20×5
500
∴P
2
=5gwtcm
−2
Case (iii) : When the block on 10cm×5cmface
Thrust= Weight of the brick =500g.wt.
Area of contact =10cm×5cm
Pressure=
Area
Thrust
=
10×5
500
P
3
=10gwtcm
−2
∴ From the above three cases, it is clear that, as the area of contact decreases, the pressure exerted increases and is greater when the brick rests on its 10cm×5cm face.