a mass m slides upward along a rough plane surface inclined at angle 023 in radians to the horiz

First, we need to consider the forces acting on the mass. There are three forces: the gravitatio

A mass m slides upward along a rough plane surface inclined...

A mass M slides upward along a rough plane surface inclined at angle Θ = 0.23 in radians to the horizontal. Initially the mass has a speed V 0 = 4.55 m/s, before it slides a distance L = 1.0 m up the incline. After sliding this distance the new speed of the mass is V 0 / 5 measured in m/s. What is the coefficient of kinetic friction μ k between the mass and the incline?

Transcript

00:03 Hello guys in this problem we have an object of mass m the object slide towards the top of an rough inclined plane and the initial speed of the object at the bottom of the inclined plane is 4 .55 meters per second after traveling for a length of one meter along the there is speed reduced to v0 over 5 that is the 1 5th of its initial speed we have to find out the coefficient of kinetic friction for that we need to apply the work energy equation and also we need to draw the free body diagram let's draw the free body diagram at an instant during the motion the weight of the mass act along the vertical downward direction and the weight equals the mass m times g a normal force n act normal to the surface and a kinetic friction force f k act towards the bottom of the inclined plane now let's take the component of the weight normal to the brain and parallel to the plane the component of the weight normal to the brain is m g cost theta and parallel to the brain is mg sine theta the displacement of the object is towards the top d the angle made by normal to the displacement is 90 degree and the angle made by the perpendicular component of the weight with the displacement is 90 degree from the work energy equation we can write the network 10 equals the change in kinetic energy the networked in equals minus of mg sine theta times the displacement along the inclined plane l minus the kinetic friction force f k times the length l equals half times v0 over 5 square minus v0 square times the mass m say this has equation well now we need to find out the kinetic friction force if we consider normal to the plane as positive y direction and towards the top as positive x then we can write the net force along y is 0 and we get the normal force equals m g cos theta therefore the kinetic friction force equals the coefficient of kinetic friction times mg cos theta so we can rewrite this equation as minus of mg sine theta times l minus the coefficient of kinetic friction mg cos theta l equals 0 .5 times the mass m times with negative sign 24 v0 square over 25 okay now we can cancel out the mass from each term and we can write the coefficient of kinetic friction times g g g g x theta l equals 12 v0 square over 25 minus g sine theta l or we get the coefficient of kinetic friction equals 12 v0 square minus 25 g sine theta l over 25 g sine theta uh sorry 25 g g g g g g g g g l, substitute the values...

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