I need some help finishing my physics homework. If you can help me I will forever be grateful to you.
1. A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 823 N.
(a) As the elevator moves up, the scale reading increases to 954 N, then decreases back to 823 N. Find the acceleration of the elevator.
(b) As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator?
(c) Using your results from parts a and b, explain which change in velocity, starting or stopping, would take the longer time.
(d) What changes would you expect in the scale readings on the ride back down?
2. You are driving a 2450.0 kg car at a constant speed of 14.0 m/s along an icy, but straight, level road. As you approach an intersection, the traffic light turns red. You slam on the brakes. Your wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 23.7 m. What is the coefficient of kinetic friction between your tires and the icy road?How do you solve physics equations?!?Student weighs 823 N (at rest) and mass = 823/g kg
1)
a) F = ma gives 954 - 823 = (823/g)a where a = acc of lift and g = 9.81
So a ~ 0.1592g ~ 1.56 ms^-2
b) F = ma gives 782 - 823 = (823/g)a where a = acc of lift and g = 9.81
So a ~ -0.0498g ~ -0.4887 ms^-2
c) The initial accn is 1.56 which is ~ 3.2 times the deceleration so the lift gains speed faster than it loses speed. This is better from the passengers' points of view.
d) The ride down would have the following normal reactions:
mg - N1 = m(1.56) down
N2 - mg = m(0.4887) and you can find N1 and N2
2) W.D by frictional force = change in KE
F. 23.7 = (1/2) mv^2 - (1/2)mu^2
F. 23.7 = 0 - (1/2)2450(14^2)
F ~ -10 130 N = (mu) mg where mu = coeff of friction.
Negative sign indicates a retarding force
so mu = 0.422 (taking g as 9.81)
