(a) As it falls 85 m, what is the work done on the raindrop by the gravitational
ID: 1783998 • Letter: #
Question
(a) As it falls 85 m, what is the work done on the raindrop by the gravitational force?
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(b) What is the work done on the raindrop by air resistance?
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2.)A block of mass
m = 3.90 kg
is pushed a distance
d = 5.90 m
along a frictionless, horizontal table by a constant applied force of magnitude
F = 16.0 N
directed at an angle
= 22.0°
below the horizontal as shown in the figure below.
(a) Determine the work done on the block by the applied force.
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(b) Determine the work done on the block by the normal force exerted by the table.
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(c) Determine the work done on the block by the gravitational force.
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(d) Determine the work done by the net force on the block.
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5.)A particle is subject to a force Fx that varies with position as shown in the following figure.
(a) Find the work done by the force on the particle as it moves from x = 0 to x = 3.00 m.
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(b) Find the work done by the force on the particle as it moves from x = 5.00 m to x = 7.00 m.
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(c) Find the work done by the force on the particle as it moves from x = 10.0 m to x = 15.0 m.
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(d) What is the total work done by the force over the distance x = 0 to x = 15.0 m?
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(a) What is its kinetic energy at this moment?
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(b) Find the net work done on the object if its velocity changes to (8.00 î+ 4.00 ) m/s.
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7.)A 2 400-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.80 m before coming into contact with the top of the beam, and it drives the beam 15.0 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
magnitude N direction ---Select--- upward downwardExplanation / Answer
1) (a) As the weight points downwards,the work done by the gravitational force is positive the same way as does the displacement of the raindrop:
W(g) = Fg x h = mgh = 3.47 * 10^-5 kg * 9.8 N/kg * 90 m = 0.031 J
= 31 mJ
(b) Now the work done by air resistance is always negative. Since the raindrop falls at constant speed it's in equilibrium, so the magnitude of air resistance is the same as the magnitude of the raindrop's weight.
Hence
W(r) = –F(r) x h = –Fg x h = –W(g) = –31 mJ.
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