The 20 kg at angle of 53⁰ in an inclined plane is realsed from rest the coefficient of friction bn the block and the inclined plane are ųs=0.3 and ųk=0.2 A) would the block move
B) if it moves what is its speed after it has descended a distance of 5m down the plane

Given :-

To Find :-

Solution :-

For figure refer to attachment .

So the block will move if the angle of the inclined plane is greater than the angle of repose . We can find it as ,

Substitute ,

Solve ,

Hence ,

Hence the block will slide down .

Now assuming that block is released from the reset , it's initial velocity will be 0m/s .

And the net force will be ,

Substitute, N = mgcos53⁰ ( see attachment)

Take m as common,

Simplify ,

Substitute the values of sin , cos and g ,

Simplify ,

Now using the Third equation of motion namely,

Substituting the respective values,

Simplify and solve for v ,

Hence the velocity after covering 5m is 8.18 m/s .

what was the answer?

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