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A body of mass m initially at rest starts sliding from top of a smooth wedge of mass M, Height h and base angle θ . The wedge moves horizontally with const. acceleration a opposite to the direction of sliding . Calculate the time required for mass m to reach the bottom and the total change of potential energy of mass m.

A  body of mass m initially at rest starts sliding from top of a smooth wedge of mass  M,

 

Height h and base angle  θ . The  wedge moves horizontally with const.

 

acceleration a  opposite to the direction of slidingCalculate the time required for mass 

 

m to reach the bottom and the total change  of potential energy of mass m.

Grade:Upto college level

1 Answers

askIITianexpert IITDelhi
8 Points
12 years ago

Obviously change in potential energy of body =work done by gravity on the system (wedge & body combined)= mgh.{as change in pot. energy of wedge is zero;it's still on the ground !!! }

For time calculation net force(& so net acceleration) on body down the incline is calculated first by changing the non-inertial(accelerated) frame into inertial one by introducing pseudo force  ma to the opposite of direction of acceleration of wedge taking body as a system with wedge as its frame in the FBD diagram.

As the direction of hor. accn is not fully clear here,i'm giving answer for both possible case.

Consider wedge having its inclined plane being left as you draw it on screen here.(hypotenus on left & height on right)

CASE 1: accn  applied to right in hor. direction.ma will be to left in hor. direction.

                  Taking x-axis along the plane  & y-axis perpendicular to it .

                  mgsinθ+macosθ=manet (force eqnin the x-direction)

or,             t=[(2h/sinθ)/(gsinθ+acosθ)]1/2 using S=ut+1/2at2 where S=h/sinθ ,a=anet , u=0(starts from rest) with condition being gcosθ>asinθ

               as in the y-direction force eqwill give mgcosθ=masinθ+N

                  normal force=0 means body is no longer in contact of wedge,that's why the above condition .

 CASE 2: accn  applied to left in hor. direction.ma will be to right in hor. direction.

                  mgsinθ-macosθ=manet (force eqnin the x-direction)

or,             t=[(2h/sinθ)/(gsinθ-acosθ)]1/2 using S=ut+1/2at2 where S=h/sinθ ,a=anet , u=0(starts from rest) with no condition.

                   as in the y-direction force eqwill give mgcosθ+masinθ=N

                  normal force can never be zero unless a is (-ve) here which imply the first case.

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