A man whose mass is 90kg stands on a scale (calibrated in Newton) on the floor o

Question

A man whose mass is 90kg stands on a scale (calibrated in Newton) on the floor of an elevator which is moving from the 100th floor to ground. It slows down from 20 m/s to 8 m/s between 50th and the 45th floor. in a time interval of 3 seconds. a) draw a diagram of the man on the elevator. b) draw the free body diagram of just the man c) use the free body diagram to calculate the apparent weight of the man during this time interval. d) between the 45th floor and 25th floor the elevator continues to travel at 8 m/s. what is the apparent weight during this interval? e) from the 25th floor to the 15th floor the elevator speeds up to 16 m/s. what is the apparent weight during this interval?

Explanation / Answer

v=u+at = > 8 = 20+3a => a = -4 m/s^2

s= 20*3 - 4*9/2 => 42 m = distance of 5 floor

c) apparent wt=mg+ma=90*(9.8+12/3)=1242 N

d)apparent wt=real wt= 98*9.8=882 N

e) height of 25th floor to the 15th floor = 2*42 = 84 m

v^2= u^2 +2as => 16^2 = 8^2 +2*84*a => a= 1.142 m/s^2

apparent wt=mg-ma= 98*(9.8-1.142)=848.484 N

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