The breaking strength of the cable used to pull a body is 40N. A body of mass 8kg is resting on a table of coefficient of friction `mu=0.20` . The maximum acceleration which can be produced by the cable is (take , `g=10ms^(-2)`)

Options

A. `6ms^(-2)`
B. `3ms^(-2)`
C. `8ms^(-2)`
D. `9ms^(-2)`

Answer

B

Explanation

`a_("max") = (T_("max") - mumg)/(m) = (40-0.2xx8xx10)/(8) = 3ms^(-2)`