12. +0/10 points | Previous Answers SerPSE9 7 P053 My Notes Ask Your An inclined

Question

12. +0/10 points | Previous Answers SerPSE9 7 P053 My Notes Ask Your An inclined plane of angle = 20.0° has a spring of force constant k = 455 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m2.57 kg is placed on the plane at a distance d 0.273 m from the spring. From this position, the block is projected downward toward the spring with speed v 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest? 7.303 Did you accidentally divide or take the inverse in your calculation? m

Explanation / Answer

Use the principle of conservation of energy. There are three forms of energy to consider:

1. Kinetic energy due to the block's motion;
2. Gravitational potential energy due to the block's height on the ramp;
3. Elastic potential energy stored in the compressed spring.

Initially all the energy is in the block, because of its motion and its height. In the end, when the block is momentarily stopped, all the energy has been transferred to the spring.

Initial energy = KE + GPE
= ½mv² + mgh
"h" is the block's height, which is Lsin ("L" = length of ramp, 0.273 meters), so:
Initial energy = ½mv² + mgLsin

The energy stored in a spring is ½kx², where "x" is the amount of compression. So:
Final energy = ½kx²

Initial energy = final energy
½mv² + mgLsin = ½kx²

.5*(2.57kg)*(.750m/s)2 + (2.57kg)* 9.8 * .273m * sin 20 =.5 x455 * x2

3.0744 = 227.5 *  x2

x=0.116 m

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