An 80 kg man stands in an elevator. What force does the floor of the elevator exert on the man if
a) the elevator is stationary?
b) the elevator accelerates upward at 2 m/s2?
c) the elevator rises with constant velocity 4 m/s?
d) while rising, the elevator decelerates at 1.5 m/s2?
e) the elevator descends with constant velocity 7 m/s?
I'm completely confused. Can someone explain ?
a. F = M*g = 80 * 9.8 = 784 N.
b. F-784 = M*a.
F-784 = 80*2 = 160.
F = 944 N.
c. F-784 = M*a.
F-784 = M*0 = 0.
F = 784 N.
d. F-784 = M*a.
F-784 = 80*(-1.5).
F-784 = -120.
F = 664 N.
Sure, I can explain how to calculate the forces in each scenario.
a) When the elevator is stationary, the floor of the elevator exerts a force equal to the weight of the man. The weight of an object can be calculated using the formula weight = mass * gravitational acceleration. In this case, the mass of the man is given as 80 kg, and the gravitational acceleration is typically taken as 9.8 m/s^2. So the force exerted by the floor of the elevator on the man would be (80 kg) * (9.8 m/s^2) = 784 N.
b) When the elevator accelerates upward at 2 m/s^2, we need to consider both the weight of the man and the additional force due to the acceleration. The total force exerted by the floor of the elevator can be calculated as the sum of these two forces. The weight of the man is the same as in part a (784 N), and the additional force due to the acceleration can be calculated as mass * acceleration. So we have (80 kg) * (2 m/s^2) = 160 N. Therefore, the total force would be 784 N + 160 N = 944 N.
c) When the elevator rises with constant velocity of 4 m/s, it means that its acceleration is 0. In this case, the force exerted by the floor of the elevator is equal to the weight of the man, which is 784 N as calculated in part a.
d) When the elevator decelerates at 1.5 m/s^2, we need to consider the direction of the deceleration. The man is still experiencing a force due to gravity (weight), but now the force exerted by the floor of the elevator will be in the opposite direction to oppose the deceleration. To calculate this force, we subtract the force due to deceleration from the weight of the man. So we have (784 N) - (80 kg) * (1.5 m/s^2) = 672 N.
e) When the elevator descends with constant velocity of 7 m/s, it means that its acceleration is 0. Similar to part c, the force exerted by the floor of the elevator would be equal to the weight of the man, which is 784 N.
In summary, the forces exerted by the floor of the elevator on the man under different conditions are:
a) 784 N (upward)
b) 944 N (upward)
c) 784 N (upward)
d) 672 N (downward)
e) 784 N (downward)