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High.
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Here in this given problem there is an inclined plane inclined at an angle theta with the horizontal.
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A block kept over it, a spring attached with the lower end of this inclined plane and being compressed under the gravitational force and this compression in the spring that is equal to d.
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If the mass of this block is m then its weight m g acting vertically downwards so its components one of the component normal to this plane incline plane because if this angle is theta this angle here that will also be theta.
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So the component of weight of the block normal to the plane that is m g cost theta and component along the inclined plane that is m g sine theta.
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So the spring will be compressed under the influence of this mg sine theta component.
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X -axis is along the inclined plane in upward direction.
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Y -axis is normal to the incline plane.
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Mass of the block that is 160 kilogram spring constant of the standard spring, ideal spring, 550 newton per meter and compression in the spring that is d.
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For the time being this is not given to us.
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Now in the first part of the problem, part a, we have to write down.
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An expression for the normal force exerted by the plane on the block and using newton's third law of motion that will be equal to m g cos theta and that is positive m g cos theta because it is along positive y -axis and here it becomes the answer for the first part of the problem that is a long positive y axis.
03:06
Then in the second part of the problem, we have to write down an expression for the forces acting on the block along the inclined plane.
03:22
And as the block is having a tendency to move up, actually there will be no net force acting on the block when the block just starts moving up...