A.) A mass m of 2.31 kg is attached to the end of a string whose length is 0.500
ID: 1979389 • Letter: A
Question
A.) A mass m of 2.31 kg is attached to the end of a string whose length is 0.500 m. The mass slides without friction on a horizontal surface as indicated in the diagram. If the string can withstand a maximum tension of 91.5 N, what is the maximum tangential speed the ball can have before the cord breaks?
B.) The mass, with a new string, is whirled in a vertical circle of the same radius about a fixed point. Find the magnitude of the tension when the mass is at the top if its speed at the top is 6.81 m/s.
A.) A mass m of 2.31 kg is attached to the end of a string whose length is 0.500 m. The mass slides without friction on a horizontal surface as indicated in the diagram. If the string can withstand a maximum tension of 91.5 N, what is the maximum tangential speed the ball can have before the cord breaks? B.) The mass, with a new string, is whirled in a vertical circle of the same radius about a fixed point. Find the magnitude of the tension when the mass is at the top if its speed at the top is 6.81 m/s.Explanation / Answer
mass m = 2.31 kglength of the string L = 0.500 m Maximum tension T = 91.5 N we know T = mv^ 2 / L from this Maximum tangential speed of the disc v =[ TL / m ] = 4.45 m / s (b). speed at top V = 6.81 m / s tension T = - mg + mv^ 2 / L = -( 2.31*9.8) +[ 2.31*6.81^2 / 0.500] = 191.61 N mass m = 2.31 kg
length of the string L = 0.500 m Maximum tension T = 91.5 N we know T = mv^ 2 / L from this Maximum tangential speed of the disc v =[ TL / m ] = 4.45 m / s (b). speed at top V = 6.81 m / s tension T = - mg + mv^ 2 / L = -( 2.31*9.8) +[ 2.31*6.81^2 / 0.500] = 191.61 N
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