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]
40 N
80 N
120 N
160 N
Explanation
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Go to new school physics book 2chapter 6(linear momentum) page 185 . It says if an elevator is going up the force which the body exerts in the ground is equal to the product of the mass and the sum of the acceleration due to gravity and the acceleration with which the elevator moves with. You can correct me if am wrong
R - W = mg
R = mg + W
(W = m×g)
R = mg + ma = m(a+g)
80N/10 = 8kg(converting weight to mass)
R = 8(5+10) = 8 × 15 = 120N
The selected answer is wrong:
The force the body exerts on the lift is equal to the force the floor of the lift exerts on the body and since force exerted on the floor by the body is M(a+g) ---->8kg(mass)*(5+10)=120
The mass M and m are supposed to be the same. Please mass remains constant. We are considering only one body with mass m.
The formular R= m(g+a) should be used instead. The answer is C. Please correct it.
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
