A man of mass 80 kg is standing in a lift, which is accelerating upwards with an acceleration of 0.6 ms-2. The size of the force between the man and the lift floor is ___N?

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Answer 1 · 2018-04-28T03:55:34

#832 \ "N"#

Use Newton's second law motion, which states that,

#F=ma#

#m# is the mass of the object in kilograms

#a# is the acceleration of the object in #"m/s"^2#

But, weight also acts in the opposite direction, so we also must subtract weight into account. The equation becomes:

#F-W=ma#

#:.F-80 \ "kg"*9.8 \ "m/s"^2=80 \ "kg"*0.6 \ "m/s"^2#

#F-784 \ "N"=48 \ "N"#

The total force needed is then:

#F=48 \ "N"+784 \ "N"#

#=832 \ "N"#