A block of mass 0.248 kg is placed on top of a light vertical spring of force co

Question

A block of mass 0.248 kg is placed on top of a light vertical spring of force constant 4980 N/m and pushed downward so that the spring is compressed 0.110 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? Enter your answer to 2 decimal places. A 1.00 kg mass slides to the right on a surface having a coefficient of kinetic friction 0.250. The object has a speed of v_i = 3.40 m/s when it makes contact with a light spring that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d. The object is then forced toward the left by the spring 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.

Explanation / Answer

1. Let the mass rise h meter from the point of release, By the law of energy conservation
=>PE(spring) at release = PE(mass) at final point
=>1/2kx^2 = mgh
=>h = kx^2/2mg
=>h = [4980 x (0.110)^2]/[2 x 0.248x 9.8]
=>h = 12.4 m

2. The kinetic energy of the mass at the moment of contact is:
KE = m×v²/2
KE = (1 kg)×(3.4 m/s)²/2
KE = 5.78 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 kg)×(9.8 m/s²)×d
Wf = (2.45 N)×d

By conservation of energy, you know that
KE = PE + Wf
(5.78 J) = (25 N/m)×d² + (2.45 N)×d
(25 N/m)×d² + (2.45 N)×d - (5.78 J) = 0

This is a quadratic equation that you can solve by using the discriminant.
25d² + 2.45d - 5.78 = 0
d = 0.4343 m
d = -0.532 m
A negative value makes no sense in this case, so the only valid solution is:
d = 0.434 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 = (5.78 J) - 2×(2.45 N)×(0.434 m)
KE = 3.65 J

The velocity at that moment is:
KE = m×v²/2
(3.65 J) = (1 kg)×v²/2
v = 2.7 m/s < - - - - - - - - - - - - - - - - - - - - - answer (b)

The distance needed to stop the mass is:
(3.65 J) = (2.45 N)×D
D = 1.48 m < - - - - - - - - - - - - - - - - - - - - - answer (c)

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