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An object moving along a horizontal track collides with and compresses a light spring (which obeys Hooke’s Law) located at the end of the tr
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An object moving along a horizontal track collides with and compresses a light spring (which obeys Hooke’s Law) located at the end of the track. The spring constant is 52.1 N/m, the mass of the object 0.250 kg and the speed of the object is 1.70 m/s immediately before the collision.
(a) Determine the spring’s maximum compression if the track is frictionless.
?? m
(b) If the track is not frictionless and has a coefficient of kinetic friction of 0.120, determine the spring’s maximum compression.
??m
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5 years
2021-07-13T03:13:04+00:00
2021-07-13T03:13:04+00:00 1 Answers
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Answers ( )
(a) As it gets compressed by a distance x, the spring does
W = – 1/2 (52.1 N/m) x ²
of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object’s displacement). By the work-energy theorem, this work is equal to the change in the object’s kinetic energy. At maximum compression x, the object’s kinetic energy is zero, so
W = ∆K
– 1/2 (52.1 N/m) x ² = 0 – 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.118 m
(b) Taking friction into account, the only difference is that more work is done on the object.
By Newton’s second law, the net vertical force on the object is
∑ F = n – mg = 0
where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives
n = mg = 2.45 N
and from this we get the magnitude of kinetic friction,
f = µn = 0.120 (2.45 N) = 0.294 N
Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:
W (friction) = – (0.294 N) x
W (spring) = – 1/2 (52.1 N/m) x ²
==> W (total) = W (friction) + W (spring)
Solve for x :
– (0.294 N) x – 1/2 (52.1 N/m) x ² = 0 – 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.112 m