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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)
 

Grade:11

2 Answers

Sandeep Bhamoo
443 Points
5 years ago
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 please approve my answer if correct
avanthika
25 Points
5 years ago
why the friction is taking upwards.Force is applied upwards so the body has a tendency to move upward.so frictional force should be downward right.I am confused with this.Plz correct where I went wrong

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