OSCAR THE OTTER IS HAVING FUN SLIDING DOWN A MUDBANK. OSCAR HAS A MASS M = 10.91

Question

OSCAR THE OTTER IS HAVING FUN SLIDING DOWN A MUDBANK. OSCAR HAS A MASS M = 10.91 KG. THE LENGTH OF THEMUD BANK IS L = 5.52 METERS. THE COEFFICIENT OF KINETICFRICTION BETWEEN OSCAR AND THE BANK IS MU = .338 THE ANGLE THE MUDBANK MAKES W.R.T. THE HORIZONTAL IS THETA = 30.1 DEGREES. AFTERWORK, OSCAR DECIDED TO CALCULATE WHAT HE COULD ABOUT THE DAYS'SFUN. HE FOUND THAT TAKING THE +X AXIS TO LIE ALONG THE BANKAND POINTING DOWN HILL SEEMED TO MAKE HIS SOLUTIONS A LITTLEEASIER. (A) CALCULATE OSCAR'S ACCELERATION DOWN HILL. (B) STARTING FROM REST AT THE TOP OF THE MUD BANK, HOW LONGWILL IT TAKE OSCAR TO SLIDE THE LENGTH OF THE HILL? (C) STARTING FROM REST FROM THE TOP OF THE HILL, WHAT ISOSCAR'S VELOCITY AS HE HITS THE WATER? OSCAR THE OTTER IS HAVING FUN SLIDING DOWN A MUDBANK. OSCAR HAS A MASS M = 10.91 KG. THE LENGTH OF THEMUD BANK IS L = 5.52 METERS. THE COEFFICIENT OF KINETICFRICTION BETWEEN OSCAR AND THE BANK IS MU = .338 THE ANGLE THE MUDBANK MAKES W.R.T. THE HORIZONTAL IS THETA = 30.1 DEGREES. AFTERWORK, OSCAR DECIDED TO CALCULATE WHAT HE COULD ABOUT THE DAYS'SFUN. HE FOUND THAT TAKING THE +X AXIS TO LIE ALONG THE BANKAND POINTING DOWN HILL SEEMED TO MAKE HIS SOLUTIONS A LITTLEEASIER. (A) CALCULATE OSCAR'S ACCELERATION DOWN HILL. (B) STARTING FROM REST AT THE TOP OF THE MUD BANK, HOW LONGWILL IT TAKE OSCAR TO SLIDE THE LENGTH OF THE HILL? (C) STARTING FROM REST FROM THE TOP OF THE HILL, WHAT ISOSCAR'S VELOCITY AS HE HITS THE WATER?

Explanation / Answer

Given that the mass of oscar is M = 10.91 kg   Length of inline is L = 5.52 m The angle of inclination is =30.1o ------------------------------------------------------------------------------------------------------------             (a)Apply Newtons second law about the length oframp                            Mg sin = Ma                                      a = g sin                                           =(9.8 m/s2) sin30.1o                                           =4.91 m/s2 (b) From the equation of motion             S = Ut + (1/2)at2             L = 0 + (1/2)at2               t = (2L / a )1/2                  = --------- sec (c) From the equation of motion the final velocity is V= U + at                                                                                    =0 +at                                                                                    =at                                                                                    =---------- m/s           (a)Apply Newtons second law about the length oframp                            Mg sin = Ma                                      a = g sin                                           =(9.8 m/s2) sin30.1o                                           =4.91 m/s2 (b) From the equation of motion             S = Ut + (1/2)at2             L = 0 + (1/2)at2               t = (2L / a )1/2                  = --------- sec (c) From the equation of motion the final velocity is V= U + at                                                                                    =0 +at                                                                                    =at                                                                                    =---------- m/s                            = --------- sec (c) From the equation of motion the final velocity is V= U + at                                                                                    =0 +at                                                                                    =at                                                                                    =---------- m/s          

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