Answer:
[tex]W_f = 148.17J[/tex]
Explanation:
During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.
Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,
[tex]N=\frac{mg}{2} = \frac{90*9.8}{2} = 441N[/tex]
The work done is given by the friction force and the distance traveled,
[tex]W_f = fd = \mu_k Nd[/tex]
Where [tex] \mu_k [/ tex] is the coefficient of kinetic friction
N is the normal force previously found d is the distance traveled,
Replacing,
[tex]W_f = (0.80)(441)(0.42)[/tex]
The thermal energy released through the work done is,
[tex]W_f = 148.17J[/tex]
The thermal energy deposited on the in the tire and the road surface is 1,452.1 J.
The given parameters;
The thermal energy deposited on the in the tire and the road surface is calculated as follows;
[tex]W = F_k d[/tex]
where;
[tex]F_k[/tex] is the frictional force
The frictional force is calculated as follows;
[tex]F_k = \mu_k F_n[/tex]
[tex]W = \mu_k F_n d\\\\W = \mu_ k (mg) d\\\\W = 0.8 \times 90 \times 9.8 \times 0.42\\\\W = 2,904.25 \ J[/tex]
The thermal energy deposited in half of the tire is calculated as;
W = 0.5 x 2,904.25 J
W = 1,452.1 J
Thus, the thermal energy deposited on the in the tire and the road surface is 1,452.1 J.
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