1.

The breaking strength of the cable used to pull a body is 40 N. A body of mass 8 kg 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=ms^(-2) )

Answer»

`6 MS^(-2)`
`3 ms^(-2)`
`8 ms^(-2)`
`8 ms^(-2)`

Solution :(B) `a_("max")=(T_("max")-mu mg)/(m)=()40-0.2xx8xx(10)/(8)=3 ms^(-2)`


Discussion

No Comment Found