Step 5: Substitute v, = 0, cancel the 0 + gy, -v? + gy, mass m, and solve for v,. v, = V 2g(y, – Y) = V 2(9.80 m/s?)(10.0 m – 5.00 m) Vr = 9.90 m/s (B) Find the diver's speed at the water's surface, y = 0. Use the same procedure as in part (a), 0 + mgy, =mv? + o v - V 2gy, - V 2(9.80 m/s?)(10.0 m) = 14.0 m/s taking y,= 0. LEARN MORE

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Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Please only answer the exercise question at the bottom
EXAMPLE 5.5
Platform Diver
GOAL Use conservation of energy to calculate the speed of a body
falling straight down in the presence of gravity.
10.0 m
KE; = 0
PE,- mgy;
PROBLEM A diver of mass m drops from a board 10.0 m above the
water's surface, as in the figure. Neglect air resistance. (a) Use
conservation of mechanical energy to find his speed 5.00 m above the
water's surface (b) Find his speed as he hits the water.
5.00 m-
STRATEGY Refer to the problem-solving strategy. Step 1: The
system consists of the diver and Earth. As the diver falls, only the
force of gravity acts on him (neglecting air drag), so the mechanical
KE, = mu,?
energy of the system is conserved, and we can use conservation of
energy for both parts (a) and (b). Step 2: Choose y = 0 for the
PE, =0
water's surface. Step 3: In part (a), y = 10.0 m and y = 5.00 m are
the points of interest, while in part (b), y = 10.0 m and y = 0 m are of interest.
SOLUTION
(A) Find the diver's speed halfway down, at y = 5.00 m.
Step 4: We write the energy
KE, + PE, = KE, + PE,
conservation equation and supply the
+ mv? + mgy;
+ mgy =
proper terms.
Step 5: Substitute v; = 0, cancel the
0 + gy, -v? + gy,
mass m, and solve for v,.
v, = V 2g(y, - v) = V 2(9.80 m/s?)(10.0 m - 5.00 m)
V = 9.90 m/s
(B) Find the diver's speed at the water's surface, y = 0.
Use the same procedure as in part (a),
taking y- 0.
0 + mgy, =mv? + 0
Vr -
V 2gy, - / 2(9.80 m/s²)(10.0 m) = 14.0 m/s
LEARN MORE
REMARKS Notice that the speed halfway down is not half the final speed. Another interesting point is
that the final answer doesn't depend on the mass. That is really a consequence of neglecting the change
in kinetic energy of Earth, which is valid when the mass of the object, the diver in this case, is much
smaller than the mass of Earth. In reality, Earth also falls towards the diver, reducing the final speed, but
the reduction is so minuscule it could never be measured.
QUESTION Qualitatively, how will the answers change if the diver takes a running dive off the end of
the board?
O Answer (a) would increase and answer (b) would decrease.
O Answer (b) would increase and answer (a) would decrease.
Both answers would decrease.
O Both answers would increase.
O Both answers would remain the same.
PRACTICE IT
Use the worked example above to help you solve this problem. A diver of mass m drops from a board 8.00
m above the water's surface, as shown in the figure. Neglect air resistance.
(a) Use conservation of mechanical energy to find his speed 4.00 m above the water's surface.
8.854
m/s
(b) Find his speed as he hits the water.
v m/s
12.522
EXERCISE
HINTS: GETTING STARTED I "M STUCK!
Use the values from PRACTICE IT to help you work this exercise. Suppose the diver vaults off the
springboard, leaving it with an initial speed of 3.04 m/s upward. Use energy conservation to find his speed
when he strikes the water.
m/s
Transcribed Image Text:EXAMPLE 5.5 Platform Diver GOAL Use conservation of energy to calculate the speed of a body falling straight down in the presence of gravity. 10.0 m KE; = 0 PE,- mgy; PROBLEM A diver of mass m drops from a board 10.0 m above the water's surface, as in the figure. Neglect air resistance. (a) Use conservation of mechanical energy to find his speed 5.00 m above the water's surface (b) Find his speed as he hits the water. 5.00 m- STRATEGY Refer to the problem-solving strategy. Step 1: The system consists of the diver and Earth. As the diver falls, only the force of gravity acts on him (neglecting air drag), so the mechanical KE, = mu,? energy of the system is conserved, and we can use conservation of energy for both parts (a) and (b). Step 2: Choose y = 0 for the PE, =0 water's surface. Step 3: In part (a), y = 10.0 m and y = 5.00 m are the points of interest, while in part (b), y = 10.0 m and y = 0 m are of interest. SOLUTION (A) Find the diver's speed halfway down, at y = 5.00 m. Step 4: We write the energy KE, + PE, = KE, + PE, conservation equation and supply the + mv? + mgy; + mgy = proper terms. Step 5: Substitute v; = 0, cancel the 0 + gy, -v? + gy, mass m, and solve for v,. v, = V 2g(y, - v) = V 2(9.80 m/s?)(10.0 m - 5.00 m) V = 9.90 m/s (B) Find the diver's speed at the water's surface, y = 0. Use the same procedure as in part (a), taking y- 0. 0 + mgy, =mv? + 0 Vr - V 2gy, - / 2(9.80 m/s²)(10.0 m) = 14.0 m/s LEARN MORE REMARKS Notice that the speed halfway down is not half the final speed. Another interesting point is that the final answer doesn't depend on the mass. That is really a consequence of neglecting the change in kinetic energy of Earth, which is valid when the mass of the object, the diver in this case, is much smaller than the mass of Earth. In reality, Earth also falls towards the diver, reducing the final speed, but the reduction is so minuscule it could never be measured. QUESTION Qualitatively, how will the answers change if the diver takes a running dive off the end of the board? O Answer (a) would increase and answer (b) would decrease. O Answer (b) would increase and answer (a) would decrease. Both answers would decrease. O Both answers would increase. O Both answers would remain the same. PRACTICE IT Use the worked example above to help you solve this problem. A diver of mass m drops from a board 8.00 m above the water's surface, as shown in the figure. Neglect air resistance. (a) Use conservation of mechanical energy to find his speed 4.00 m above the water's surface. 8.854 m/s (b) Find his speed as he hits the water. v m/s 12.522 EXERCISE HINTS: GETTING STARTED I "M STUCK! Use the values from PRACTICE IT to help you work this exercise. Suppose the diver vaults off the springboard, leaving it with an initial speed of 3.04 m/s upward. Use energy conservation to find his speed when he strikes the water. m/s
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