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A 2000kg suv accelerates from rest at a rate of 3.00m/s^2. The total amount of force resisting its motion 1500N. How much force is applied to the suvs tires by the ground to make it accelerate? The SUV now drives at a constant 18.0m/s. The total amount of force resisting it's motion is 2500 N. Now how much force is applied to the SUV's tires by the ground?

Respuesta :

As we know by Newton's law

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

[tex]F - F_f = ma[/tex]

[tex]F = F_f + ma[/tex]

here we have

[tex]F_f = 1500 N[/tex]

[tex]a = 3.0 m/s^2[/tex]

[tex]F = 1500 + (2000)(3.0) [/tex]

[tex]F = 7500 N[/tex]

so tyre apply net force of 7500 N in forward direction

After some time the SUV is travelling with constant speed

now in this case acceleration is ZERO

so now we will have

[tex]F_{net} = 0 = F - F_f[/tex]

[tex]F = F_f[/tex]

[tex]F = 2500 N[/tex]


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