Question

When a 3.40-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.60 cm. (a) If the 3.40-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it? 1.15 cm (b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position? 1.0256 Correct: Your answer is correct. J

Answer 1

a).1.147cm

b).1.0256 J

Step-by-step explanation:

Using law of force and applications Law Hooke

`F=k*d`

`d=(F)/(k)`

But first have to know force and constant in initial conditions

`F=3.4kg*9.8(m)/(s^(2)) \\F=33.32N\\k=(F)/(d)\\ d=2.6cm(1m)/(100cm)=0.026m=26x10^(-3)m`

`k=(33.23N)/(26x10^(-3))=1281.53 (N)/(m)`

a).

`d=(F)/(k)`

`d=(1.5kg*9.8(m)/(s^(2)))/(1281.53 (N)/(m))=(14.7N)/(1281.53 (N)/(m))\\  d=0.0147m=11.47x10^(-3)m`

`d=11.47x10^(-3)m*(100cm)/(1m) \\d=1.147cm`

b).

Wokr is the force done so using the equation

`d=4cm*(1m)/(100cm)=0.04m`

`Ew=(1)/(2)*k*d^(2)\\Ew=(1)/(2)*1281.53N*(0.04m)^(2)  \\Ew=1.025N*m^(2) \\Ew=1.025 J`