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A 6.0 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant, horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m.  And. V = 3.16 m/s

Respuesta :

Answer:

F= MASS *ACCELERATION

12=6*a

a=2 m/sec^2

v^2-u^2=2*a*s

v^2=2*2*3

v=sqrt(12)

v=3.5 m/sec

work done by force= change in kinetic energy

F*S=0.5*M*U^2-0.5*M*V^2

12*3=0 0.5*6*V^2

V=3.46 M/SEC

Explanation:

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