Elastic potential energy is equal to the force times the distance of movement. Elastic potential energy = force x distance of displacement. Because the force is = spring constant x displacement, then the Elastic potential energy = spring constant x displacement squared.
Answer:
a) x = (0.0114 ± 0.0001) in
, b) the number of decks is 5
Explanation:
a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated
x = d / 52
x = 0.590 / 52
x = 0.011346 in
Let's look for uncertainty
Δx = dx /dd Δd
Δx = 1/52 Δd
Δx = 1/52 0.005
Δx = 0.0001 in
The result of the calculation is
x = (0.0114 ± 0.0001) in
b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is
#_cards = n 52
The formula for thickness is
x = d / n 52
Uncertainty
Δx = 1 / n 52 Δd
n = 1/52 Δd / Δx
n = 1/52 0.005 / 0.00002
n = 4.8
Since the number of decks must be an integer the number of decks is 5
Answer:
If the particle is an electron 
If the particle is a proton, 
Explanation:
Initial speed at the origin, 
to +ve x-axis
The particle crosses the x-axis at , x = 1.5 cm = 0.015 m
The particle can either be an electron or a proton:
Mass of an electron, 
Mass of a proton, 
The electric field intensity along the positive y axis
, can be given by the formula:

If the particle is an electron:



If the particle is a proton:



Answer:
a) F = 680 N, b) W = 215 .4 J
, c) F = 1278.4 N
Explanation:
a) Hooke's law is
F = k x
To find the displacement (x) let's use the elastic energy equation
= ½ k x²
k = 2
/ x²
k = 2 85.0 / 0.250²
k = 2720 N / m
We replace and look for elastic force
F = 2720 0.250
F = 680 N
b) The definition of work is
W = ΔEm
W =
- 
W = ½ k (
² - x₀²)
The final distance
= 0.250 +0.220
= 0.4750 m
We calculate the work
W = ½ 2720 (0.47² - 0.25²)
W = 215 .4 J
We calculate the strength
F = k 
F = 2720 0.470
F = 1278.4 N