Solving Spring Truck Problem 2: Compute Work, Speed, if Force is Conservative

[TyFelmingham]

2. A spring from the truck is found not to conform to Hooke’s law. The force law for this spring is
found to be:
F = – k1x – k2 x^2
where k1 = 52.8 N.m-1; k2 = 38.4 N.m-2.
a) Compute the work required to stretch the spring from x = 0.500 m to x = 1.00 m.
b) With one end of the spring fixed, a particle of mass 2.17 kg is attached to the other end of the
spring when it is extended by an amount x = 1.00 m. If the particle is then released from rest,
compute its speed at the instant the spring has returned to the configuration in which the extension
is x = 0.500 m.
c) Is the force exerted by the spring conservative or nonconservative? Explain your answer.

Ok, for a)

sub in the numbers for x = 1m and you get -91.2 N
sub in the numbers for x= .5m and you get - 36 N

So force is 127.2 N when the spring is stretched from .5 to 1m?

As W = F*x, do i just plug in x=.5m and F = 127.2 N?

Cheers

 

[KingOfTwilight]

In a and b. Calculate the energy in x = 0,5 and x = 1,00. You can't do a with the simple W = Fs because F is changeable.

 

[TyFelmingham]

In a and b. Calculate the energy in x = 0,5 and x = 1,00. You can't do a with the simple W = Fs because F is changeable.

So you have to integrate?

 

[Naeem]

Yeah, I think so,

Work done = Integral F.dx : Limits of Integraton are from x = 0.5 to x = 1.00

 

[TyFelmingham]

Yeah, I think so,

Work done = Integral F.dx : Limits of Integraton are from x = 0.5 to x = 1.00

Cheers