A box of mass 5kg is placed on an inclined plane of angle 35 degree.A force of 10N is applied parallel to the plane in a direction up the plane is applied to the box.The coefficient of sliding friction is 0.07.Determine the acceleration of the box(ans -3.05m/s^2 down the plane)

Question

Arun · Answer

Sin(35)=0.57, Cos(35)=0.819. Now making fbd ,downward force (due to gravity) parallel to plane=5g•sin(35) ≈28.6 Upward force (given) parallel to plane =10. Upward force (frictional) = (0.07)5g•cos(35) ≈2.867. Hence net downward force =28.6–10–2.867 ≈ 15.153 now acceleration =15.153/5 ≈3.031m/s² Regards Arun (askIITians forum expert)

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A box of mass 5kg is placed on an inclined plane of angle 35 degree.A force of 10N is applied parallel to the plane in a direction up the plane is applied to the box.The coefficient of sliding friction is 0.07.Determine the acceleration of the box(ans -3.05m/s^2 down the plane)

A box of mass 5kg is placed on an inclined plane of angle 35 degree.A force of 10N is applied parallel to the plane in a direction up the plane is applied to the box.The coefficient of sliding friction is 0.07.Determine the acceleration of the box(ans -3.05m/s^2 down the plane)

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A box of mass 5kg is placed on an inclined plane of angle 35 degree.A force of 10N is applied parallel to the plane in a direction up the plane is applied to the box.The coefficient of sliding friction is 0.07.Determine the acceleration of the box(ans -3.05m/s^2 down the plane)

A box of mass 5kg is placed on an inclined plane of angle 35 degree.A force of 10N is applied parallel to the plane in a direction up the plane is applied to the box.The coefficient of sliding friction is 0.07.Determine the acceleration of the box(ans -3.05m/s^2 down the plane)

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