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A trunk of mass 22 kg is on the floor. The trunk has a very small initial speed. The acceleration of gravity is 9.8 m/s 2 .What constant horizontal force pushing the trunk is required to give it a velocity of 10 m/s in 20 s if the coefficient of sliding friction between the trunk and the floor is 0.51? Answer in units of N.

Respuesta :

amhugueth

Answer:

[tex]F=120.96N[/tex]

Explanation:

Given the information from the exercise we need to use Newton's Laws to solve it

If we analyze the y-axis

∑[tex]F_{y}=N-W=0[/tex]

[tex]N=W=m*g[/tex]

Since the trunk is moving in the x-axis and its velocity is changing with time, its acceleration is:

[tex]a=\frac{10m/s}{20s}=0.5m/s^2[/tex]

Knowing that we can analyze the forces acting in the horizontal direction

∑[tex]F_{x}=F-fr=m*a[/tex]

[tex]F=m*a+fr[/tex]

[tex]F=m*a+uN=m*a+u*m*g[/tex]

[tex]F=(22kg)(0.5m/s^2)+(0.51)(22kg)(9.8m/s^2)=120.96N[/tex]

So, the force required to push the trunk is 120.96 N

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