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Gala2k [10]
2 years ago
11

A 2.0 kg pendulum has an initial total energy of 20 J. Calculate the energy lost as heat if the pendulum is 0.10 m high and is t

raveling at 4.0 m/s.
A) 2.0J
B) 18J
C) 14J
D)16J
Physics
1 answer:
Maurinko [17]2 years ago
4 0

The correct answer is (A) 2.0 J

Total energy of the pendulum is the sum of its kinetic and potential energy. At the instant of time, when the pendulum is at a height <em>h</em> and has a speed <em>v, </em>Its energy is given by,

E=mgh+\frac{1}{2} mv^2

Substitute 2.0 kg for <em>m</em>, the mass of the pendulum, 9.81 m/s² for <em>g</em>, the acceleration due to gravity, 0.10 m for <em>h and 4.0 m/s for </em>v<em>.</em>

E=mgh+\frac{1}{2} mv^2\\ =(2.0kg)(9.81m/s^2)(0.10 m)+\frac{1}{2}(2.0kg)(4.0m/s)^2\\ =17.962J

The pendulum has an initial energy of 20 J. the energy lost is given by,

\Delta E=(20J)-(17.962J)\\ =2.038J=2.0J

Thus, the energy lost by the pendulum is (A) 2.0 J

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a)

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2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

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Substituting, we find the net force on the bicyclist:

F_{net}=(60)(3.1)=186 N

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

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The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

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Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

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The centripetal acceleration for the fourth radius; 8.0 m

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The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

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