a block of 10kg mass is placed on a rough inclined surface as shown in figure. The acceleration of the block will be
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Answered by
23
Given that the body experiences both static and kinetic friction we can get the resultant frictional force acting on the body.
The resultant coefficient of frictional force is :
1 - 0.8 = 0.2
The forces acting on the object are as follows :
Mg = 10 × 10 = 100
Normal force = mg Cos Ф = 100 Cos 30 = 86.60N
The parallel force = mg Sin Ф = 100 Sin 30 = 50 N
The frictional force = ωNormal force where ω is the coefficient of friction.
= 0.2 × 86.6 = 17.32 N
The resultant force on the object is :
50 N - 17.32 N = 32.68 N
Acceleration = F/mass
Acceleration = 32.68/10 = 3.268 m/s²
pragyasharma24:
If the angle of inclination is greater than angle of repose,tan^{-1}\mu, the acceleration of block will be g sin\theta-\mu cos\theta, and if the angle is lesser that that, the acceleration is zero. Hence, the acceleration of the block will be zero.
Answered by
44
Find the value of frictional force
=uN (where N= mgcostheta)
=10g√3/2
.
.
Now find accelerating force in downward direction
=mgsintheta
=10g/2
.
.it is clear thst frictional force is more than acc. Force that is tje reason that the block wont move
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And thus acc. Is 0
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