A body weighing 80N stands in an elevator that ... - JAMB Physics 2002 Question
A body weighing 80N stands in an elevator that is about to move. The force exerted by floor on the body as the elevator moves upward with an acceleration of 5 ms-2 is
[g = 10ms-2]
[g = 10ms-2]
A
40 N
B
80 N
C
120 N
D
160 N
correct option: d
Let M = mass of elevator
m = mass of the body
g = acceleration of free fall
a = upward acceleration of the elevator (5m/ss)
when the elevator is at rest, the force it exerts on the body is equal to the weight of the body. And from Newtons third law, Mg = mg
However as the elevator ascends the net upward force F = Mg - mg = (M + m)a.
Thus, if the weight of the body = 80 N, then 88 its mass, m = 80/10 = 8kg.
Therefore M x 10 - 8 x 10 = (M + 8) 5
10M - 80 = 5M + 40
10M - 5M = 40 x 80
5M = 120 => M = 120/5 = 24kg.
Thus, net upward force = force exerted by the elevator floor on the body = (M x m)a
= (24 x 8) x5
= 32 x 5
= 160 N
m = mass of the body
g = acceleration of free fall
a = upward acceleration of the elevator (5m/ss)
when the elevator is at rest, the force it exerts on the body is equal to the weight of the body. And from Newtons third law, Mg = mg
However as the elevator ascends the net upward force F = Mg - mg = (M + m)a.
Thus, if the weight of the body = 80 N, then 88 its mass, m = 80/10 = 8kg.
Therefore M x 10 - 8 x 10 = (M + 8) 5
10M - 80 = 5M + 40
10M - 5M = 40 x 80
5M = 120 => M = 120/5 = 24kg.
Thus, net upward force = force exerted by the elevator floor on the body = (M x m)a
= (24 x 8) x5
= 32 x 5
= 160 N
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