Map Due Date Sapling Learning Late subn A 2970 kg block of w

Map Due Date Sapling Learning Late subn A 2.970 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is 0.555 and the coefficient of kinetic friction is 0.305. At time t = 0, a force F 9.94 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times Points Pos Grade Cat Descriptio Policies: t=0 t>0 Number You can c You can v You have There is rn Consider the same situation, but this time the external force Fis 20.1 N. Again state the force of friction acting on the block at the following times: D eTextbod t=0 t>0 O Help Wit Number o Web Help Technical PreviousCheck Answer NextExit

Solution

maximum friction force. f = us m g

f = 0.555 x 2.970 x 9.81 = 16.17 N

Applied force is less than max. friction force hence it will not move.

t = 0 : f = 9.94 N ......Ans

t > 0 f = 9.94 N ........Ans

When F = 20.1 N

at t = 0 , f = 16.17 N .........Ans

after that it will start moving hence kinetic friction will act on it.

f = 0.305 x 2.970 x 9.81

f = 8.89 N .............Ans