A) 0.51
B) 0.41
C) 0.71
D) 0.61.
Correct Answer: D
Solution :
We know that while a cyclist moving with a speed v takes a sharp turn on a circular track of radius r, the coefficient of friction is given by \[\mu =\tan \theta =\frac{{{v}^{2}}}{rg}\] Here \[v=4.9\text{ }m/s\] \[r=4\text{ }m\] and \[g=9.8\text{ }m/{{s}^{2}}\] \[\therefore \] \[\mu =\frac{4.9\times 4.9}{4\times 9.8}=0.61\]You need to login to perform this action.
You will be redirected in
3 sec