A motorcycle of mass 160 kg accelerates from rest to 53 m/s in 9 seconds. Ignore air resistance. Assuming there's no slipping between the wheels and the pavement of the road.

Required:
a. What is the average horizontal component of the force that the road exerts on the wheels (total force on all two wheels, not the force on one wheel)?
b. How far does the motorcycle travel in 9 seconds?
c. In the point-particle analysis of this situation, what is the work done by this force?
d. For the real system, how much work is done by the force of the road on the wheels?

Question

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Respuesta :

okpalawalter8 okpalawalter8 · Answer 1 · 2021-07-18T03:19:28+02:00

Answer:

a) [tex]F=940.8N[/tex]

b) [tex]S=234.14m[/tex]

c) [tex]W=2.2*10^5J[/tex]

d) [tex]W=0[/tex]

Explanation:

Mass [tex]m=160kg[/tex]

Velocity [tex]v=53m/s[/tex]

Time [tex]t=9seconds[/tex]

a)

Generally the Newton's equation for motion is mathematically given by

[tex]a=\frac{v}{t}[/tex]

[tex]a=\frac{53}{9}[/tex]

[tex]a=5.9m/s^2[/tex]

Therefore

F=ma

[tex]F=160*5.88[/tex]

[tex]F=940.8N[/tex]

b)

Generally the Newton's equation for motion is mathematically given by

[tex]S=0.5at^2[/tex]

[tex]S=0.5*5.9*9^2[/tex]

[tex]S=234.14m[/tex]

c)

Generally the Newton's equation for work done is mathematically given by

[tex]W=Fd[/tex]

[tex]W=940.8*238.14[/tex]

[tex]W=2.2*10^5J[/tex]

d)

Generally the Newton's equation for work done by the force of the road on the wheels is mathematically given by

[tex]W=Fdcos\theta[/tex]

[tex]W=0[/tex]