A block of mass 2kg slides down an incline plane of inclination 30. The coefficient of friction between block and plan is 0.5. The contact force between block and plan is?

Question

Arun · Answer

Dear Dhruv Contact force is the resultant of Frictional force and normal reaction force. The frictional force can be calculated as = 0.5 x mgcos theta. = 0.5 x 2 x 10 x root 3 / 2 normal reaction force is Mgcos theta. = 2 x 10 x root 3 / 2

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A block of mass 2kg slides down an incline plane of inclination 30. The coefficient of friction between block and plan is 0.5. The contact force between block and plan is?

A block of mass 2kg slides down an incline plane of inclination 30. The coefficient of friction between block and plan is 0.5. The contact force between block and plan is?

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A block of mass 2kg slides down an incline plane of inclination 30. The coefficient of friction between block and plan is 0.5. The contact force between block and plan is?

A block of mass 2kg slides down an incline plane of inclination 30. The coefficient of friction between block and plan is 0.5. The contact force between block and plan is?

1 Answers

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