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Q. A block is projected upwards on an inclined plane of inclination 37° along the line...
Q. A block is projected upwards on an inclined plane of inclination 37° along the line of greatest slope with velocity of 5 m/s. The coefficeint of friction between the block and the surface is `mu` = 0.5 . The block first stops at a distance of ________ from starting point.
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The block is projected upwards on an inclined plane along the line of greatest slope with velocity 5 m/s. The coefficient of friction between the block and the surface is `mu = 0.5` .
There is gravitational force of attraction acting on the block that accelerates it in the vertically downwards direction. Take this as 10 m/s^2. This acceleration can be divided into a component parallel to the inclined plane and a component perpendicular to the plane. The former is 10*sin 37 and the latter is 10*cos 37. The component perpendicular to the plane provides a normal force of m*10*cos 37 and the corresponding frictional force is m*10*cos 37*0.5. The deceleration due to this force of the block is 10*cos 37*0.5
As the block moves up, the total deceleration of the block is 10*sin 37 + 10*cos 37*0.5. Use the relation v^2 - u^2 = 2*a*s where u is the initial velocity, v is the final velocity, a is the acceleration and s is the distance traveled.
The block stops at a distance s, where 0 - 25 = 2*(-10*sin 37 -10*cos 37*0.5)*s
=> s = 1.2485 `~~` 1.25 m
The correct answer is option A.
Posted by justaguide on July 7, 2013 at 3:29 AM (Answer #1)
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