InterviewSolution
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InterviewSolution
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Using the diagram from the previous example, assurme that m=2 kg, M= 10 kg, and the coefficient of kinetic friction between the small block and the tabletop is 0.5. What is the acceleration of the blocks ? |
Answer» SOLUTION :Once again, draw a free-body diagram for each object. Notice that the only difference between these diagrams and the ones in the previous EXAMPLE is the INCLUSION of friction, `F_(f)` that acrs on the block on the table. As before, we have TWO equations that contain two unknowns `(a and F_(T))`. `F_(T)-F_(f)= ma(1)` `Mg- F_(T)=Ma(2)` We add the equations (thereby eliminating `F_(T))` and solve for a. Notice that, by DEFINITION, `F_(f)= mu F_(N)` and from the free body diagram for m, we see that `F_(N)= mg`, so `F_(f)= mu mg` : `Mg-F_(f)= ma+Ma` `Mg- mu mg= a(m+M)` `(M-mu m)/(m+M) g= a` Or, using our shorter method :  Substituting in the numberical values given for m, M, and `mu_(k)`, we find that `a= (3)/(4)g` ( or `7.5 m//s^(2))`. |
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