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An elevator accelerates upward at 1.2 m/s 2 . the acceleration of gravity is 9.8 m/s 2 . what is the upward force exerted by the floor of the elevator on a(n) 76 kg passenger? answer in units of n.

Answer :

We can find the force by using the following formula;
N = ma +  mg
Fa = ma = 76 x 1.2 = 91.2
Fg = mg = 76 x 9.8 = 744.8
N = 91.2 + 744.8 = 836
So, the force is 836 N.
onyebuchinnaji

The upward force exerted on the passenger is 836 N.

The given parameters;

  • acceleration of the elevator, a = 1.2 m/s²
  • acceleration due to gravity, g = 9.8 m/s²
  • mass of the object, m = 76 kg

The upward force on the object can be determined by applying Newton's second law of motion.

Since the elevator is ascending upwards, the force on the elevator will become greater and the passenger feels a heavier weight.

∑F = ma + mg

F = m(a + g)

F = 76(1.2 + 9.8)

F = 836 N.

Thus, the upward force exerted on the passenger is 836 N.

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