Hooke's Law. If the distance is 27 cm when the weight is 2 kg, what is the distance when the weight is 5 kg?

Answer 1

we need a constant

d = Cw
27 = 2C
C = 13.5

d = 13.5 * 5
d = 67.5cm

don't stretch your mind too far.

Answer 2

"sic tensio huc vis"
as u know
F = k*x
where x is the elongation of the spring... becouse i think u are talking about a spring...
then u can determin k from the first "situation"
2 Kg * 9.81 m/s^2 = k * 27 cm
so u have
k = F/x that lead to
k = 19.62 N / 0.27 m = 72.67 N/m
and u have the final elungation x2...
x2 = 5 kg*9.81 m/s^2 / 72.67 N/m = 0.67 m = 67 cm

(1 kg is a misure of mass, not weight)

Answer 3

Since Hooke's law supposes a linear relationship between distance and force, you can use an equivalent fraction to solve this.

27/2 = x/5

Cross multiply

135 = 2x

x = 67.5

Answer 4

2:27::5:x or 2/27 = 5/x

x = 27 X 5/2 = 135/2 = 67.5 cm

The above assumes that the spring is in elastic range.

Answer 5

if distance here is extension then
extension is directly proportional to weight
therefore
2.5 times weight => 2.5 times extension
therefore extension = 2.5*27 = 67.5 cm
all weights must be under limits for hookes law to be valid
your wire must be of lenght about few hundreds of metres

Answer 6

The spring constant is 13.5 cm/kg.
Just multiply by any weight to get the corresponding extension.

The assumption is that the spring constant is valid for the range of weights considered.

Answer 7

D = k W
27 = 2 k
k = 13.5
D = 13.5 x 5 cm
D = 67.5 cm

Answer 8

the distance is 65cm