A 1.50 kg snowball is fired from a cliff 13.5 m high. The snowball\'s initial ve
ID: 1968585 • Letter: A
Question
A 1.50 kg snowball is fired from a cliff 13.5 m high. The snowball's initial velocity of 14.0 m/s, directed 41.0° above the horizontal.(a) How much work is done on the snowball by its weight during its flight to the ground below the cliff?
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(b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight?
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(c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?
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Explanation / Answer
Mass of the snow ball m =1.50 kg Height of the cliff h = 13.5 m Initial velocity u = 14.0 m/s Angle = 41.0o ------------------------------------------------ (a) Work is done on the snowball by its weight during its flight to the ground below the cliff is W= mgh = (1.50 kg)(9.8 m/s2 )(13.5 m) = 198.45 J ----------------------------------------------------------------------------- ( b) Since, the force of gravity is conservative so, the change inpotential energy U = -W = -(1.50 kg)(9.8 m/s2 )(13.5 m) = -198.45 J ----------------------------------------------------------------------------- (c) The potential energy with respect to the top of the cliff is U = -W = -(1.50 kg)(9.8 m/s2 )(13.5 m) = -198.45 J Mass of the snow ball m =1.50 kg Height of the cliff h = 13.5 m Initial velocity u = 14.0 m/s Angle = 41.0o ------------------------------------------------ (a) Work is done on the snowball by its weight during its flight to the ground below the cliff is W= mgh = (1.50 kg)(9.8 m/s2 )(13.5 m) = 198.45 J ----------------------------------------------------------------------------- ( b) Since, the force of gravity is conservative so, the change inpotential energy U = -W = -(1.50 kg)(9.8 m/s2 )(13.5 m) = -198.45 J ----------------------------------------------------------------------------- (c) The potential energy with respect to the top of the cliff is U = -W = -(1.50 kg)(9.8 m/s2 )(13.5 m) = -198.45 J ------------------------------------------------ (a) Work is done on the snowball by its weight during its flight to the ground below the cliff is W= mgh = (1.50 kg)(9.8 m/s2 )(13.5 m) = 198.45 J ----------------------------------------------------------------------------- = 198.45 J ----------------------------------------------------------------------------- ( b) Since, the force of gravity is conservative so, the change inpotential energy U = -W = -(1.50 kg)(9.8 m/s2 )(13.5 m) = -198.45 J ----------------------------------------------------------------------------- (c) The potential energy with respect to the top of the cliff is U = -W = -(1.50 kg)(9.8 m/s2 )(13.5 m) = -198.45 J = -198.45 J ----------------------------------------------------------------------------- (c) The potential energy with respect to the top of the cliff is U = -W = -(1.50 kg)(9.8 m/s2 )(13.5 m) = -198.45 J = -198.45 J ----------------------------------------------------------------------------- (c) The potential energy with respect to the top of the cliff is U = -W = -(1.50 kg)(9.8 m/s2 )(13.5 m) = -198.45 J = -(1.50 kg)(9.8 m/s2 )(13.5 m) = -198.45 J = -198.45 J = -198.45 JRelated Questions
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