A 1.50-kg object slides to the right on a surface having a coefficient of kineti
ID: 1357000 • Letter: A
Question
A 1.50-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 3.50 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e).
(a) Find the distance of compression d.
m
(b) Find the speed v at the unstretched position when the object is moving to the left (Figure d).
How much energy is stored in the spring just before the object makes contact with the spring? How much energy is stored in the spring after the object leaves the spring? Do you need to consider the spring potential energy in working this part? m/s
(c) Find the distance D where the object comes to rest.
m
Explanation / Answer
The kinetic energy of the mass at the moment of contact is:
KE = m×v²/2
KE = (1.5 kg)×(3.5 m/s)²/2
KE = 5.25 J
When the mass comes to rest at point d, the potential energy stored in the spring is:
PE = k×d²/2
PE = (50 N/m)×d²/2
PE = (25 N/m)×d²
When the mass comes to rest at point d, the work done by friction is:
Wf = µ×m×g×d
Wf = 0.25×(1.5 kg)×(9.8 m/s²)×d
Wf = (3.675 N)×d
By conservation of energy, you know that
KE = PE + Wf
(5.25 J) = (25 N/m)×d² + (3.675 N)×d
(25 N/m)×d² + (3.675 N)×d - (5.25 J) = 0
This is a quadratic equation that you can solve by using the discriminant.
25d² + 3.675d - 5.25 = 0
d = 0.390 m < - - - - - - - - - - - - - - - - - - - - - answer (a)
When back at equilibrium, the spring/mass system only has kinetic energy. It will be the same as the initial kinetic energy, minus the work done by friction over a distance of 2 times d.
KE = (5.25 J) - 2×(3.675 N)×(0.390 m)
KE = 2.38 J
The velocity at that moment is:
KE = m×v²/2
(2.38 J) = (1.5 kg)×v²/2
v = 1.78 m/s < - - - - - - - - - - - - - - - - - - - - - answer (b)
The distance needed to stop the mass is:
(2.38 J) = (3.675 N)×D
D = 0.64 m < - - - - - - - - - - - - - - - - - - - - - answer (c)
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