Question

A fisherman is fishing from a bridge and is using a "49.0-N test line." In other words, the line will sustain a maximum force of 49.0 N without breaking. What is the weight of the heaviest fish that can be pulled up vertically, when the line is reeled in (a) at constant speed and (b) with an acceleration whose magnitude is 2.93 m/s2?

Answer 1

Solution:

Maximum force = F = 49 N .

a) When it is moving at constant speed, acceleration = 0 .And the maixmum force itself is the weight of the heaviest fish .

Weight of the heaviest fish that can be pulled at constant speed = 49 N

Mass = 49 /9.8 = 5 kg = Maximum mass of the fish .

b) SInce there is an upward acceleration , the acceleration is now greater than 9.8 m/s/s

F_N - mg = ma

=> F_N = mg + ma = m(g+a)

= 5( 9.8 + 2.93) = 63.7 N = weight during acceleration .

Mass when there is 2.93 m/s/s acceleration = 49 /(9.8+2.93) = 3.85 kg .