2.) Find the total work done on 12-N block if there is no friction 3.) Find the

Question

2.) Find the total work done on 12-N block if there is no friction

3.) Find the total work done on 20-N block if us=.5 and uk= .325 between the table and the block

4.) Find the total work done on 12-N block if us=.5 and uk=.325 between the table and the block

Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure (Figure 1) ). The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward. 1.) Find the total work done on 20-N block if there is no friction 2.) Find the total work done on 12-N block if there is no friction 3.) Find the total work done on 20-N block if us=.5 and uk= .325 between the table and the block 4.) Find the total work done on 12-N block if us=.5 and uk=.325 between the table and the block

Explanation / Answer

let
w1 = m1g = 20N
w2 = m2g = 12N
s = 0.75m

the work done on w1 is
W1 = m1as = w1as/g
so you need to find a

note their is only one force moving the system ie w2
newton's law
a = ?F/?m = w2/(m1+m2) = w2g/(w1+w2) = 12g/32 = 3g/8
then
W1 = w1as/g = 3w1s/8 = 3*20*0.75/8 = 5.625J
______________________________________...
Just out of interest
The work done on w2 is
W2 = m2as = w2as/g = 3w2s/8 = 3*12*0.75/8 = 3.375J
so the total work is
W = W1 + W2 = 9J
which is equal to the change in potential energy, as expected
,.,.,.

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