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a particle of mass m=375 g is launched with velocity of A =4 moves with a length AB=2.5m under the action of tractive force F=3.46 N making an angle 30. during its motion between A and B which is subjected to a frictional force f=1.5 N
calculate V of B by applying the kinetic energy theorem​

Respuesta :

Numenius

Answer:

The final speed is 5.78 m/s.

Explanation:

mass, m = 375 g = 0.375 kg

initial velocity, u = 4 m/s

Distance, s = 2.5 m

Angle, A = 30 degree

Force, F = 1.5 N

let the final velocity is v.

Use the work energy theorem

Work done = change in kinetic energy

[tex]W= 0.5 m(v^2 - u^2)\\\\F s cos A= 0.5 m (v^2 - u^2)\\\\1.5\times 2.5\times cos30= 0.5\times 0.375\times (v^2 - 16)\\\\v = 5.78 m/s[/tex]

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