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A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 450 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude (in N) 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?

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opudodennis

Answer:

1185 N

Explanation:

From Newton’s second law of motion,  

F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force

Net force acting on motorcycle [tex]F_{net}[/tex] is given by  [tex]F_{net}=F-f[/tex]

Where F is force acting on motorcycle and f is frictional force

Substituting F-f for [tex]F_{net}[/tex]

[tex]F_{net}=ma[/tex] hence  ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided

245*3= F- 450

F=245*3 +450= 1185 N

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