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The mass of the heavier block is 24 kg and the mass of the lighter blocks is 12 kg and the magnitude of the force of the connecting string on the smaller block is 19 N. Assume: g = 9.8 m/s 2 and the horizontal surface on which the objects slide is frictionless. 12 kg 24 kg 19 N F Determine the force F. Answer in units of N.

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Answer:

[tex]F=57N[/tex]

Explanation:

Recall that in this case, friction is not taken into account. Thus, Using the free body diagram, according to Newton's second law, we have:

For the lighter block:

[tex]\sum F_x=T=m_1a[/tex]

For the heavier block:

[tex]\sum F_x=F-T=m_2a(1)[/tex]

The acceleration is the same for both blocks:

[tex]a=\frac{T}{m_1}(2)[/tex]

So, replacing (2) in (1):

[tex]F-T=m_2\frac{T}{m_1}\\F=T(1+\frac{m_2}{m_1})\\F=19N(1+\frac{24kg}{12kg})\\F=57N[/tex]

Answer:

the force F is 56.992 N

Explanation:

the solution is in the attached Word file

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