We got this worksheet for physics, and i cannot figure this question out. The worksheet included the answers, and we had to do the work for it and see if we get the same answer. I do not know how to do this.
PLEASE INCLUDE YOUR WORK AND EXPLAIN-- Thank you!!!
--------------------------------------…
# The drop of Doom's cage has a mass of 300 kg. It takes 4.08 seconds to fall the total distance of 20.0 meters. CALCULATE:
A) Potential Energy at the top---The answer is (5.89x10^4 J)-- I got this no problem!!
B) Speed at bottom. answer: (19.8 m/s) i figured this out-- no problem
C) If the Cage stops in a distance of 19.0 meters, what force did the brakes apply? answer: (3.10x10^3 N) PROBLEM!!!!!!! HOW DID THEY GET THE ANSWER. I DONT UNDERSTAND
PLEASE SHOW WORK FOR THIS
D) How long will it take the cage to come to a stop? answer: (1.92 s)-- PROBLEM!!! I DONT GET THIS ONE EITHER.. PLEASE SHOW WORK FOR THIS ONE TOO.
THE QUESTION I HAD TROUBLE WITH are (((C and D)))-- i dont know how to do these-- so please show work and explain!!!
THANK YOU!!!!!!
PLEASE INCLUDE YOUR WORK AND EXPLAIN-- Thank you!!!
--------------------------------------…
# The drop of Doom's cage has a mass of 300 kg. It takes 4.08 seconds to fall the total distance of 20.0 meters. CALCULATE:
A) Potential Energy at the top---The answer is (5.89x10^4 J)-- I got this no problem!!
B) Speed at bottom. answer: (19.8 m/s) i figured this out-- no problem
C) If the Cage stops in a distance of 19.0 meters, what force did the brakes apply? answer: (3.10x10^3 N) PROBLEM!!!!!!! HOW DID THEY GET THE ANSWER. I DONT UNDERSTAND
PLEASE SHOW WORK FOR THIS
D) How long will it take the cage to come to a stop? answer: (1.92 s)-- PROBLEM!!! I DONT GET THIS ONE EITHER.. PLEASE SHOW WORK FOR THIS ONE TOO.
THE QUESTION I HAD TROUBLE WITH are (((C and D)))-- i dont know how to do these-- so please show work and explain!!!
THANK YOU!!!!!!
-
I'm not familiar with "Cage of Doom" but I must assume that it is a carnival ride which drops vertically 20 meters and rapidly changes direction to a horizontal track where the braking occurs. I have to assume this to get the numbers to work out to the given answers..
C If the track is now horizontal, no more energy is added to the system by gravity.
so all the energy collected in the fall must be dissipated as heat of friction in braking.
E = Fd
E = 5.89x10^4 J from our first answer gives us
5.89x10^4 = F(19)
F = 3.1 x 10^3
D) from F = ma
our deceleration rate is
3.1 x 10^3 = 300a
a = 10.33 m/s/s
from our displacement/acceleration equations
y = (1/2)at^2
19 = (1/2)(10.33)t^2
t = 1.92 seconds
C If the track is now horizontal, no more energy is added to the system by gravity.
so all the energy collected in the fall must be dissipated as heat of friction in braking.
E = Fd
E = 5.89x10^4 J from our first answer gives us
5.89x10^4 = F(19)
F = 3.1 x 10^3
D) from F = ma
our deceleration rate is
3.1 x 10^3 = 300a
a = 10.33 m/s/s
from our displacement/acceleration equations
y = (1/2)at^2
19 = (1/2)(10.33)t^2
t = 1.92 seconds
-
hi,
work done = gravitational potential energy:
f*d=5.89x10^4
f=5.89x10^4/19
f=3100
d) f=ma
a=3100/300
=31/3
use equations of motion
s=19
u=19.8
v=0
a=-31/3
t=?
v=u+at
t=v-u/a
t=0-19.8/-(31/3)
1.92 s
please reply if you cant follow anything done here and ill be gald to help.
work done = gravitational potential energy:
f*d=5.89x10^4
f=5.89x10^4/19
f=3100
d) f=ma
a=3100/300
=31/3
use equations of motion
s=19
u=19.8
v=0
a=-31/3
t=?
v=u+at
t=v-u/a
t=0-19.8/-(31/3)
1.92 s
please reply if you cant follow anything done here and ill be gald to help.