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A block of Mass M is moving with a velocity v on straight surface. What is the shortest distance and shortest time inwhich the block can be stopped ifU is coefficient of frictionV()022ug ug02ug ugU22()2M 9' ugNone of the above |
| Answer» FORCE of friction=μN=μmgTherefore RETARDATION = μmg/m=μgFrom,or,from v=u+ator, t=v/μg | |