Jale2468
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As respresented in the diagram below a constant 15-newton force, f is applied to a 2.5-kilogram box,accelerating the box to the right at 2.0 meters per second squared across a rough horizontal surface

Explain the formulas and steps you use

As respresented in the diagram below a constant 15newton force f is applied to a 25kilogram boxaccelerating the box to the right at 20 meters per second squared class=

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#51, 52

Net force on the box is given by sum of all forces on the box

As per Newton's II law we can say

[tex]F_{net} = ma[/tex]

it is product of mass and acceleration of the object

[tex]F_{net} = 2.5kg * 2m/s^2[/tex]

given that

mass = 2.5 kg

acceleration = 2 m/s^2

[tex]F_{net} = 5 kg m/s^2[/tex]

here we know that

[tex]1 kg m/s^2 = 1 N[/tex]

so we can say that

[tex]F_{net} = 5 N[/tex]

#53

Now we can say there are two forces acting on the box

1. applied force which is given as 15 N in forward direction

2. friction force on the box in opposite direction.

[tex]F_a - F_f = F_{net}[/tex]

[tex]15 - F_f = 5[/tex]

[tex]F_f = 15 - 5 = 10 N[/tex]

so friction force on this box will be 10 N