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What is its velocity and position 15 seconds after it is launched? A block is of

ID: 1416523 • Letter: W

Question

What is its velocity and position 15 seconds after it is launched? A block is of mass 4kg is moving at 200m/s and is projected up an inline plane with Theta = 30 degrees. Mu k = 0.2. How high does it go up before coming to a stop. How much heat energy is created in going to the top? When it comes back down what is its speed when it reaches the bottom? Three 5kg balls are located above with distance between A and B = 0.4m and B and C 0.8m. What are the magnitude and direction (with respect to positive x axis) of the let gravitational force of on sphere B due to balls A and C?

Explanation / Answer

1) Here, 1/2 * 4 * 200 * 200   = 4 * 9.8 * sin30 * x   + 0.2 * 4 * 9.8 * cos30 * x

=>   x   = 3031.49 m         (distance along incline)

=> height above surface =   3031.49 * sin30   =   1515.74 m

=> heat energy created =   0.2 * 4 * 9.8 * cos30 * 3031.49

                                          = 20582.72 J

=>   1/2 * 4 * v2 = 4 * 9.8 * sin30 * 3031.49   - 0.2 * 4 * 9.8 * cos30 * 3031.49

=> speed when it reach bottom , v = 139.34 m/sec

2)   net force along x axis   = 6.67 * 10-11 * 5 * 5/0.82

                                         = 2.605 * 10-9 N

      net force along y axis   = 6.67 * 10-11 * 5 * 5/0.42

                                         = 10.421 * 10-9 N

=> resultant force   =    10.741 * 10-9 N

=> direction   = tan-1(10.421/2.605)

                      =   76 degree

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