Finding acceleration in spring situations

AI Thread Summary
A duck with a mass of 40 kg is pulled by a spring with a spring constant of 900 N/m, stretching 0.4 m. The calculations reveal the spring force to be 360 N, while the frictional force, calculated using a coefficient of kinetic friction (μk) of 0.6, is 236 N. The net force acting on the duck is determined to be 124 N after accounting for friction. Using Newton's second law, the acceleration of the duck is calculated to be 3.1 m/s². The discussion highlights the importance of correctly identifying forces and their directions in solving physics problems.
bunk66
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Homework Statement


A duck of mass 40kg is being pulled by a spring with k = 900N/m. If μk = 0.6 between the duck and the road, find the acceleration of the duck given that the spring stretches 0.4m.

Homework Equations


Fspring = kx, where k = spring constant and x = stretch of spring
Fnet = ma

The Attempt at a Solution


Fspring = kx
F = 900N/m(0.6m)
F = 540N

That's about as far as I got before not knowing what to do next.
 
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bunk66 said:
F = 900N/m(0.4m)
F = 540N
Those are two different forces, so give them different names.
In which directions do they act?
How do they relate to the acceleration of the duck?
 
haruspex said:
Those are two different forces, so give them different names.
In which directions do they act?
How do they relate to the acceleration of the duck?
What exactly do you mean that they are two different forces? In the section that you quoted, isn't there only one force being calculated: the spring force?

And p.s, I made a typing error. Where it says 0.4, it should be 0.6
 
bunk66 said:
I made a typing error. Where it says 0.4, it should be 0.6
OK, that's what made me think you were referring to two different forces.
So, what is the frictional force acting on the duck (direction and magnitude)?
 
haruspex said:
OK, that's what made me think you were referring to two different forces.
So, what is the frictional force acting on the duck (direction and magnitude)?
I mixed some of the numbers apparently.

Fspring = kx
F = 900N/m(0.4m)
F = 360N

40kg x 9.8 = 392N

Ff = μk(FN)
Ff = 0.6(392N)
Ff = 236N

Fnet = Fa - Ff
Fnet = 360N - 236N
Fnet = 124N

Fnet = ma
Fnet/m = a
124N / 40kg = a
3.1 m/s2 = a

Evidently that was a lot easier than I made it out to be :redface:
 
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