Practice 5E: | 1 | 2 |
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**Conservation of mechanical energy - by Matt
Henderson , 2003**

**1. A bird is flying with a speed of 18.0 m/s over water
when it accidentally drops a 2.00 kg fish. If the altitude of the bird is 5.40m
and friction is disregarded, what is the speed of the fish when it hits the
water?**

Here's what you know, v1 = 18 m/s and m = 2.00 kg and h = 5.4 m since KE=PE then KE + PE = PE2 + KE2 right. KE =

^{1}/_{2}mv^{2 }and PE = mgh so^{ 1}/_{2}mv^{2 +}mgh = mgh(2) +^{ 1}/_{2}mv^{2 (2). }We know that h1 = 5.4 and h2 = 0 (thats when it hits the water) also that m = 2.00 kg and g = 9.8 m/s^2 and v1 = 18.0 m/s, but we don't know v2 = ? . So we plug the numbers in and solve (1/2)(2)(18^{2}) + (2)(9.8.)(5.4) = (1/2)(2)(V2^{2}) + (2)(9.8)(0)

^{v = 20.73 m/s }

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**2. A 755 N diver drops from a board 10.0 m above the
water's surface. Find the diver's speed 5.00 m above the water's surface. Then
find the diver's speed just before striking the water. **

F = 755N, and h = 10.0 m and a = 9.8 m/s^2 and m =? and v = ?

To find the mass use F = ma 755 = (m)(9.8) and m = .77 kg then use the formula KE + PE = PE2 + KE2 which is

^{1}/_{2}mv^{2 }+ mgh = mgh(2) +^{ 1}/_{2}mv^{2 (2) }to solve for the velocity. v1 = 0 and h1 = 10 and v2 = ? and h2 = 0 (1/2)(.77)(0^2) + (.77)(9.8.)(10) = (1/2)(.77)(V2^2) + (.77)(9.8)(0)

^{v = 14 m/s }

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**3. If the diver in item 2 leaves the board with an
initial upward speed of 2.00 m/s, find the diver's speed when striking the
water. **

v(i) = 2.00m/s First we need to figure out high the diver went off the board. We know that KE=PE and we know m = .77 kg and g = 9.8 m/s^2 In order to find the height we plug it into the formula

^{1}/_{2}mv^{2}^{ = }^{mgh }

^{(1/2)(.77)(2^2) = (.77)(9.8)(h1)}h1 = .20468 m Now we add that to 10 m. This gets us the total height the diver fell from. 10 + .20468 = 10.20468No that we know h1 we can find the final velocity by using KE=PE again. which is

^{1}/_{2}mv^{2}^{ = }^{mgh }

^{(1/2)(.77)(v}^{2}^{^2) = (.77)(9.8)(10.2)}

v2 = 14.14 m/s

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**4. An Olympic runner leaps over a hurdle. If the runner's
initial vertical speed is 2.2 m/s, how much will the runner's center of
mass be raised during the jump?**

We know v =2.2 m/s the center of mass is the same as the height, so we use KE=PE which is

^{1}/_{2}mv^{2 }= mghthe mass cancel out so we are left with (1/2)(v^2) = (g)(h) plug it in (1/2)(2.2^2) = (9.8)(h)

^{h = .247 m }

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**5. A pendulum bob is released from some initial height
such that the speed of the bob at the bottom of the swing is 1.9 m/s. What is
the initial height of the bob? **

v = 1.9 m/s h = ? same as the last problem use 1/2mv = mgh

^{(1/2)(1.9^2) = (9.8)(h)}

^{h = .18 m }

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