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An 8 kg block is placed at the top of a plane inclined by 30° with a coefficient of kinetic friction of 0.1. What is the block’s acceleration down the ramp?
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- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
An 8 kgblock is placed at the top of a plane inclined by 30°with a...
- To obtain the acceleration, we must obtain the net force. A free-body diagram will show the forces acting on the block is motionless in the direction perpendicular to the plane but is present in the direction parallel into the plane.
- The two forces acting on the block are mg sin θ, the component of gravitational force acting downwards along the plane, and Ff the force of kinetic friction acting directly against motion and upwards along the plane.
- Frictional force is the product of the normal force and the coefficient of kinetic friction: Ff = μkFN
- Since the axis that runs perpendicular to the plane shares the angle θ with respect to the direction of gravitational force, we determine the value of the gravitational force down the slope of the plane’s surface to be mg sin θ and the value of the gravitational force perpendicular to the plane (also the value of the normal force) to be mg cos θ. Therefore, the frictional force is Ff = μk mg cos θ.
- The net force down the ramp is the difference between the two forces:
Dividing Fnet by m gives us net acceleration:
Replacing the variables with numbers we get:
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