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Question

a block of mass m is placed on a horizontal surface. If the block is pulled by applying a force of magnitude F=5mg at an angle θ=370 with horizontal as shown in Fig.6.29, find the acceleration of the block at the given instant.
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Solution

Three forces act on the block (a) mg (b) F=5mg (c) N
Resolving the forces along x-and y-axis, we have equation of motion
Fx=max
4mg=max
This gives ax=4g.......(i)
Fy=may
3mgmg+N=may
2mg+N=may.......(iii)
Let ay=0. Then N=2mg
The negative result signifies that N will be directed down (opposite to the assumed direction), but the ground cannot pull the block down. Hence, the block will lose contact with the ground.
or N=0
Hence, from (ii), ay=2g
Hence, the net acceleration, |a|=a2x+a2y
Where ax=4g from Eq. (i) and ay=2g from Eq. (ii)
This gives a=a2x+a2y=(4g)2+(20)2=25g
Analysis of Newton's laws of motion in connected bodies : Problems based on normal reaction

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