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?
Please help ASAP!! Step by step along with the equations to use
Well, well, well, we've got ourselves an elevator conundrum! Let's see if I can help you out in true Clown Bot style.
a) When the elevator is stationary, the floor of the elevator exerts a normal force equal to the man's weight. That means it's simply m*g, where m is the mass of the man (80 kg) and g is the acceleration due to gravity (9.8 m/s^2). So the force the floor exerts on the man is 80 kg * 9.8 m/s^2 = 784 N.
b) Now, when the elevator accelerates upward at 2 m/s^2, the floor has to do a little extra work to keep that man from floating off. In this case, we have to consider the net force acting on the man. The equation to use here is F_net = m*a, where F_net is the net force, m is the mass (80 kg), and a is the acceleration (2 m/s^2). Since the floor exerts the normal force, opposite to the direction of motion, we can write the equation as F_net = N - m*g = m*a. Solving for N, we get N = m*(a + g) = 80 kg * (2 m/s^2 + 9.8 m/s^2) = 784 N + 160 N = 944 N.
c) Ah, when the elevator rises with a constant velocity of 4 m/s, the acceleration is zero, my friend! So the floor only has to exert enough force to counteract the man's weight. Therefore, the force the floor exerts on the man is the same as in part a), which is 784 N.
d) When the elevator decelerates at 1.5 m/s^2 while still rising, things get a little funky. We can still use the equation F_net = m*a, but in this case, the acceleration is negative. So now, F_net = N - m*g = m*(-a) since a is now -1.5 m/s^2. Solving for N, we get N = m*(g - a) = 80 kg * (9.8 m/s^2 - 1.5 m/s^2) = 80 kg * 8.3 m/s^2 = 664 N.
e) Lastly, when the elevator descends with a constant velocity of 7 m/s, we're back to no acceleration. So, the force the floor exerts on the man is once again the same as in part a), 784 N.
And voila, there you have it! The forces exerted by the elevator floor on the man in various scenarios. Hope that helped, and remember, laughter is the best elevator music!
Sure! I'd be happy to help you with this problem step-by-step.
a) When the elevator is stationary, the man is not experiencing any acceleration. In this case, the force exerted by the floor of the elevator is equal to the weight of the man, which can be calculated using the equation:
Force = mass * acceleration due to gravity
Force = 80 kg * 9.8 m/s^2
b) When the elevator accelerates upward at 2 m/s^2, the man experiences an upward force in addition to his weight. The force exerted by the floor of the elevator can be calculated by adding the force due to gravity and the force due to acceleration:
Force = mass * (acceleration due to gravity + acceleration of the elevator)
Force = 80 kg * (9.8 m/s^2 + 2 m/s^2)
c) When the elevator rises with constant velocity of 4 m/s, the man is not experiencing any acceleration. In this case, the force exerted by the floor of the elevator is equal to the weight of the man, which can be calculated using the same equation as in part a.
Force = 80 kg * 9.8 m/s^2
d) While rising, if the elevator decelerates at 1.5 m/s^2, the man experiences a smaller upward force. Similar to part b, the force exerted by the floor of the elevator can be calculated by subtracting the force due to deceleration from the force due to gravity:
Force = mass * (acceleration due to gravity - deceleration of the elevator)
Force = 80 kg * (9.8 m/s^2 - 1.5 m/s^2)
e) When the elevator descends with constant velocity of 7 m/s, the man is not experiencing any acceleration. In this case, the force exerted by the floor of the elevator is equal to the weight of the man, which can be calculated using the same equation as in part a.
Force = 80 kg * 9.8 m/s^2
I hope this helps! Let me know if you have any further questions.
Sure! Let's go through each scenario step by step.
a) When the elevator is stationary, the man experiences the force due to gravity, which is his weight. The weight of the man can be calculated using the formula:
Weight = mass × acceleration due to gravity = m × g
Since the weight is equal to the force exerted by the floor, the answer is:
Force = Weight = m × g
where m is the mass of the man (80 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, in this case:
Force = 80 kg × 9.8 m/s^2 = 784 Newtons
b) When the elevator accelerates upward at 2 m/s^2, the man experiences an additional force due to this acceleration. The force exerted by the floor can be calculated by adding the force due to gravity (weight) to the force due to acceleration:
Force = Weight + mass × acceleration
Weight = m × g = 80 kg × 9.8 m/s^2 = 784 Newtons
Acceleration force = m × a = 80 kg × 2 m/s^2 = 160 Newtons
So, in this case:
Force = 784 + 160 = 944 Newtons
c) When the elevator rises with a constant velocity of 4 m/s, there is no acceleration. In this case, the force exerted by the floor is equal to the force due to gravity (weight), as there are no additional forces acting on the man:
Force = Weight = m × g = 80 kg × 9.8 m/s^2 = 784 Newtons
d) While rising, if the elevator decelerates at 1.5 m/s^2, the force exerted by the floor can be calculated by subtracting the force due to acceleration from the force due to gravity:
Force = Weight - mass × acceleration
Weight = m × g = 80 kg × 9.8 m/s^2 = 784 Newtons
Deceleration force = m × a = 80 kg × 1.5 m/s^2 = 120 Newtons
So, in this case:
Force = 784 - 120 = 664 Newtons
e) Finally, when the elevator descends with a constant velocity of 7 m/s, there is no acceleration. Similar to case c), the force exerted by the floor is equal to the force due to gravity (weight):
Force = Weight = m × g = 80 kg × 9.8 m/s^2 = 784 Newtons
I hope this helps! Let me know if you have any further questions.
F-mg = m a
a) mg
b) m(g+2)
c) mg
d) m(g-1.5)
e) mg