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`        A 1kg object slide to the right on a surface having a coeffient of of kinetic friction 0.250. The object has a speed of vi=3m/s when it makes contact with a light spring that has a force constant of 50N/M. The object comes to rest after the spring has been compressed a distance d. The object is then forced towards the left by the spring and continues to move in that direction beyond the springs unstretched position. Finally the object comes to rest a distance D to the left of the unstretched spring.Find a. The distance of compression d, b. The speed v at the unstretched position when the object is moving to the left, and c. The distance D where the object comes to rest`
5 months ago

Vikas TU
10041 Points
```							Dear student a),(initial kinetic energy) = (potential energy of compressed spring) + (work done by friction) mv^2/2 = kd^2/2 + μmgd (1 kg)(3 m/s)^2 / 2 = (50 N/m)(d^2) / 2 + (0.250)(1 kg)(9.81 m/s^2)(d) 0 = 25d^2 + 2.4525d - 4.5 Solving quadratic equation: d = (-2.4525 + sqrt[(2.4525)^2 - 4(25)(-4.5)]) / (2*25) d = 0.378 m b) (potential energy of compressed spring) = (kinetic energy when spring is relaxed) + (more work done by friction) kd^2/2 = mu^2/2 + μmgd (50 N/m)(0.378 m)^2 / 2 = (1 kg)(u^2) / 2 + (0.250)(1 kg)(9.81 m/s^2)(0.378 m) u = 2.30 m/s c) (kinetic energy when spring is relaxed) = (even more work done by friction) mu^2 / 2 = μmgD u^2 / 2 = μgD (2.30 m/s)^2 / 2 = (0.250)(9.81 m/s^2)(D) D = 2.16 m
```
5 months ago
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