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Grade: 12
        
consider the inclined plane with an angle 0f 67 degree. The spring constant has a value of 1850 N/m. A 10kg block is pushed against the spring compressing it to 40 cm. The block is then released and the spring expands pushing the block up the plane. the coefficient of kinetic friction between the block and the plane is 0.5. what is the normal force and frictional force acting on the system?
2 years ago

Answers : (1)

Deepak Kumar
31 Points
							
Angle of inclined plane = 67degree.
Spring constant = K = 1850N/m.
Mass of block = M=10kg.
Compressed length = x = 40cm = 0.40m.
Coefficent of Kinetic friction = \upsilon = 0.5.
Spring force = Kx = 1850*0.40 = 740N (towards the inclined plane).
Component of gravitational force towards inclined plane = -Mgsin(670) = -10*9.8*sin(670) = -90.2095N.
Maximum friction force = \upsilonMgcos(670) = 0.5*10*9.8*cos(670) = 19.145825N.
Since spring force is greater than component of gravitational force so friction force is -19.145825N(negative sign indicate that force is opposite to spring force.
Normal force = Mgcos(670) = 38.29165N.
 
2 years ago
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