A m 1 = 4.91-kg block is placed on top of a m 2 = 10.0-kg block (figure). A hori

Question

A

m1 = 4.91-kg

block is placed on top of a

m2 = 10.0-kg

block (figure). A horizontal force of

F = 45.0 N

is applied to the 10.0-kg block, and the 4.91-kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0.182.

(a) Draw a free-body diagram for each block and identify the action–reaction forces between the blocks. (Submit a file with a maximum size of 1 MB.)

(b) Determine the tension in the string. (Enter the magnitude only.)

N

Determine the magnitude of the acceleration of the 10.0-kg block.

Explanation / Answer

Given that

m1 =4.91kg

m2 =10kg

F =45N

uk =0.182

Now for the top block from the free body diagram

T<----- --------->Ff

        |N

a)

Now the horizontal force for mass ma is given by

T =ukN =ukmg =0.182*4.91kg*9.81 =8.766N

b)

Now fro the block m2 the vertical force is

N =Ff =(10+4.91)(9.81) =146.118N

Now the horizontal forces are

FH =F -Ff-Ffb

ma =45-8.766N-(0.182)(14.91)(9.81) =9.613

10a =9.613

Then the acceleration (a) =9.613/10 =0.9613m/s2

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