A body initially at rest is acted upon by a constant force of 18lb for 5 sec, after which an opposite force of 12lb id applied. In what additional time will the body come to rest
initial force F1 =18 lb=80 N
initialaccelertn a1 = "80\\over m"
final velocity before second force applied will be
"v=u +at"
"v=0+{80\\over m}t ={80\\over m}\\times 5={400\\over m}"
now opposite force F2 =12lb=53.37 N is applied
and final velocity is 0.
"0={400\\over m}-{53.37\\over m}t"
we get t=7.5 sec.
Comments
Leave a comment