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A powerful motorcycle can produce an acceleration of 3.50 m/s2 while traveling at 90.0 km/h. at that speed the forces resisting motion, including friction and air resistance, total 400 n. (air resistance is analogous to air friction. it always opposes the motion of an object.) what is the magnitude of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?

Respuesta :

meerkat18

Using Newton’s 2nd law, we can calculate for the force (F) required to give a mass 255kg an acceleration of 3.5 m/s^2: 
F= ma

F= 255kg*3.5m/s^2 = 892.5 N 

892.5 was the force needed for the motorcycle alone. Taking into account the total resistance of 400N: 
total F = 892.5 N + 400 N
total F= 1292.5 N

Answer:

F = 1257.5 N

Explanation:

As per Newton's II law we know that net force on the system of mass is given as

[tex]F = ma[/tex]

[tex]F - F_{air} = ma[/tex]

here we know that

[tex]F_{air} = 400 N[/tex]

m = 245 kg

[tex]a = 3.50 m/s^2[/tex]

now we have

[tex]F - 400 = 245\times 3.50[/tex]

so we will have

[tex]F = 400 + (245)(3.50)[/tex]

[tex]F = 1257.5 N[/tex]

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