To find the force exerted by the elevator floor on a 52 kg passenger when the elevator is accelerating downwards at \( 2.4 , \text{m/s}^2 \), we can use Newton's second law of motion and consider the forces acting on the passenger.
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Calculate the weight of the passenger (force due to gravity): \[ F_g = m \cdot g \] where:
- \( m = 52 , \text{kg} \) (mass of the passenger)
- \( g = 9.8 , \text{m/s}^2 \) (acceleration due to gravity)
\[ F_g = 52 , \text{kg} \cdot 9.8 , \text{m/s}^2 = 509.6 , \text{N} \]
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Calculate the net force acting on the passenger: Since the elevator accelerates downwards at \( 2.4 , \text{m/s}^2 \), the net force can be calculated as: \[ F_{\text{net}} = m \cdot a \] where \( a = 2.4 , \text{m/s}^2 \).
\[ F_{\text{net}} = 52 , \text{kg} \cdot 2.4 , \text{m/s}^2 = 124.8 , \text{N} \]
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Determine the force exerted by the elevator floor (normal force): The normal force \( F_N \) is given by: \[ F_N = F_g - F_{\text{net}} \]
Thus, substituting in the values: \[ F_N = 509.6 , \text{N} - 124.8 , \text{N} = 384.8 , \text{N} \]
Therefore, the force that the elevator floor has on the 52 kg passenger is approximately 384.8 N.
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