- #1
TyFelmingham
- 9
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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
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
