Asked by AP Physics
Hello
given that I person 65 kg takes 4.5 s to climb a flight of stairs that is 5.5 m
my question is well I was asked to solve for the work and
W=fd
and in this case f is equal to the normal force correct? Just take the opposite of mg???????
Also I'm asked to solve for PE and sense PE also equal mgh...
is PE the same as work in this situation????
a similar problem a person is running forwards on a flat surface or 5 m long with a mass of 65 kg and takes 1.75 s starting from rest
I'm asked to solve for the work done
sense F=ma
how do I find the acceleration???
My last question
I'm given a spring that has a mass of 100 g and a constant of 22 N/m with an A of 5 cm
I belive the A is amplitude
and I'm asked for the Total Energy which i don't know how to do
the max velocity also have no idea how to do
and the max force which i don't know how to do
THANK You
given that I person 65 kg takes 4.5 s to climb a flight of stairs that is 5.5 m
my question is well I was asked to solve for the work and
W=fd
and in this case f is equal to the normal force correct? Just take the opposite of mg???????
Also I'm asked to solve for PE and sense PE also equal mgh...
is PE the same as work in this situation????
a similar problem a person is running forwards on a flat surface or 5 m long with a mass of 65 kg and takes 1.75 s starting from rest
I'm asked to solve for the work done
sense F=ma
how do I find the acceleration???
My last question
I'm given a spring that has a mass of 100 g and a constant of 22 N/m with an A of 5 cm
I belive the A is amplitude
and I'm asked for the Total Energy which i don't know how to do
the max velocity also have no idea how to do
and the max force which i don't know how to do
THANK You
Answers
Answered by
Damon
As for the stair problem, you have the right idea. The work, mgh, goes into increasing the potential energy.
Answered by
Damon
accelerates starting from rest
d = (1/2) a t^2
5 meters = (1/2) a (1.75)^2
solve for a, the acceleration
d = (1/2) a t^2
5 meters = (1/2) a (1.75)^2
solve for a, the acceleration
Answered by
Damon
m = 100 grams = .1 kg
k = 22 N/m
A = 5 cm = .05 m
y = .05 sin w t
w = sqrt (k/m)
you could solve for w
then maximum stretch = .05 m and velocity is zero there so
Maximum potential energy = maximum energy = (1/2) k .05^2 (This is because the kinetic energy and velocity are zero when the spring is stretched maximum)
Max force = k A
to get maximum velocity
(1/2) m v^2 = max kinetic energy = max total energy = max potential energy which we know
k = 22 N/m
A = 5 cm = .05 m
y = .05 sin w t
w = sqrt (k/m)
you could solve for w
then maximum stretch = .05 m and velocity is zero there so
Maximum potential energy = maximum energy = (1/2) k .05^2 (This is because the kinetic energy and velocity are zero when the spring is stretched maximum)
Max force = k A
to get maximum velocity
(1/2) m v^2 = max kinetic energy = max total energy = max potential energy which we know
Answered by
AP Physics
Thank You
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