Let's draw the free body diagram for the first case,
Suppose, "F_1=4N" and "F_2=6N" ,thus from the free body diagram the net force acting on "m=3.2Kg" is
"F_{net}=F_1+F_2\\\\\n\\implies F_{net}=4+6=10N"Now, from Newton's second law,
"F_{net}=ma"where, "a" is the acceleration of the object "m." Hence,
"10=(3.2)a\\implies a=\\frac{25}{8}ms^{-2}"in the rightward direction.
Now, suppose the force acting on the mass "m" is in opposite direction,thus free body diagram will be,
Hence, net force acting on "m" is
"F_{net}=F_2-F_1\\implies F_{net}=2N"Again by applying Newton's second law we get,
in the direction of "F_2" force i.e rightward direction.
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