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# A force F=20N is applied to a block at rest as shown.After the block has moved a distance of 8m to the right, the direction of horizontal component of the force F is reversed.Find distance travelled before block stops after the F is reversed. Asked by m.nilu 14th August 2018, 10:17 PM Figure shows the forces acting on 2 kg block.

In vertical direction, we have,  N + F sin37 = 2g .........(1)
where N is normal reaction force, we get N = 2g- Fsin37 = 2×9.8 - 20×.6 = 7.6 Newton

hence in horizontal direction, we have, F×cos37 - μ×N = m×a .........(2)
where μ is the coefficent of friction and a is acceleration

we substitute values in eqn.(2) and get a = 7.05 m/s2

Starting from rest, if the block moves 8 m with acceleration a, then speed v of block after travelling 8 m is given by ,

v = after moving 8m distance, if the horizontal component reversed, the applied force gives retardation to the movement of block.

when the horizontal component is reversed but still the block is moving implies that
frictional force and applied force both are in same direction. hence retardation ar is given by

F×cos37 + μ×N = m×a.....................(3)
Hence ar = 8.95 m/s2

Distance travelled before stopping :- S = u2 / (2ar ) = (10.62×10.62)/(2×8.95) = 6.3 m

Answered by Expert 18th August 2018, 5:19 PM
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Tags: distance