A) 0.2
B) 0.8
C) 0.4
D) 0.6
Correct Answer: B
Solution :
The resultant force on the body when force F is applied and the force of friction is\[f\]and its acceleration is \[5m/{{s}^{2}}\] is given by \[m{{a}_{1}}=F-f\] \[10\times 5=F-f\] ?(i) Similarly, if the acceleration is \[18\,m/{{s}^{2}}\]and the applied force on the body is doubled IF and force of friction is\[f,\] then resultant force on the body is given by \[m{{a}_{2}}=2F-f\] \[10\times 18=2F-f\] Solving Eqs. (i) and (ii) \[f=80\,N\] Also, the force of friction \[f=\mu R\](where \[\mu \]is the coefficient offriction) or \[\mu =\frac{f}{R}=\frac{f}{mg}\] \[\mu =\frac{80}{10\times 10}=0.8\]You need to login to perform this action.
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