Question

A block of mass 250gm is given an initial speed of 3m/s half a plane inclined at an angle of 30degree with the horizontal.Coefficient of friction√3/6.After 1.5 sec how far is the block from original position​

Answer 1

Explanation:

C

36.6 m

Stopping distance S=

2a

u

2

a=gsinθ+μgcosθ

=9.8×

2

1

+0.2×9.8×

2

3

=6.6m/s

2

Now S=

2a

u

2

=

2×6.6

22×22

=36.6m

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