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A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a speed of 8.0 m/swhen it pass

arlik [135]

Answer:

a) $v_{60^{o}} =4.98 m/s$

b) $\theta_{max}=79.34^{o}$

Explanation:

This problem can be solved by doing an energy analysis on the given situation. So the very first thing we can do in order to solve this is to draw a diagram of the situation. (see attached picture)

So, in an energy analysis, basically you will always have the same amount of energy in any position of the pendulum. (This is in ideal conditions) So in this case:

$K_{lowest}+U_{lowest}=K_{60^{o}}+U_{60^{0}}$

where K is the kinetic energy and U is the potential energy.

We know the potential energy at the lowest of its trajectory will be zero because it will have a relative height of zero. So the equation simplifies to:

$K_{lowest}=K_{60^{o}}+U_{60^{0}}$

So now, we can substitute the respective equations for kinetic and potential energy so we get:

$\frac{1}{2}mv_{lowest}^{2}=\frac{1}{2}mv_{60^{o}}^{2}+mgh_{60^{o}}$

we can divide both sides of the equation into the mass of the pendulum so we get:

$\frac{1}{2}v_{lowest}^{2}=\frac{1}{2}v_{60^{o}}^{2}+gh_{60^{o}}$

and we can multiply both sides of the equation by 2 to get:

$v_{lowest}^{2}=v_{60^{o}}^{2}+2gh_{60^{o}}$

so we can solve this for $v_{60^{o}}$ . So we get:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

so we just need to find the height of the stone when the pendulum is at a 60 degree angle from the vertical. We can do this with the cos function. First, we find the vertical distance from the axis of the pendulum to the height of the stone when the angle is 60°. We will call this distance y. So:

$cos \theta = \frac{y}{4m}$

so we solve for y to get:

$y = 4cos \theta$

so we substitute the angle to get:

y=4cos 60°

y=2 m

so now we can find the height of the stone when the angle is 60°

$h_{60^{o}}=4m-2m$

$h_{60^{o}}=2m$

So now we can substitute the data in the velocity equation we got before:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

$v_{60^{o}} = \sqrt{(8 m/s)^{2}-2(9.81 m/s^{2})(2m)}$

so

$v_{60^{o}}=4.98 m/s$

b) For part b, we can do an energy analysis again to figure out what the height of the stone is at its maximum height, so we get.

$K_{lowest}+U_{lowest}=K_{max}+U_{max}$

In this case, we know that $U_{lowest}$ will be zero and $K_{max}$ will be zero as well since at the maximum point, the velocity will be zero.

So this simplifies our equation.

$K_{lowest} =U_{max}$

And now we substitute for the respective kinetic energy and potential energy equations.

$\frac{1}{2}mv_{lowest}^{2}=mgh_{max}$

again, we can divide both sides of the equation into the mass, so we get:

$\frac{1}{2}v_{lowest}^{2}=gh_{max}$

and solve for the height:

$h_{max}=\frac{v_{lowest}^{2}}{2g}$

and substitute:

$h_{max}=\frac{(8m/s)^{2}}{2(9.81 m/s^{2})}$

to get:

$h_{max}=3.26m$

This way we can find the distance between the axis and the maximum height to determine the angle of the pendulum about the vertical.

y=4-3.26 = 0.74m

next, we can use the cos function to find the max angle with the vertical.

$cos \theta_{max}= \frac{0.74}{4}$

$\theta_{max}=cos^{-1}(\frac{0.74}{4})$

so we get:

$\theta_{max}=79.34^{o}$

5 0

3 years ago

How much would it cost to operate five incandescent light bulbs for the months of April and May, collectively, given that each b

const2013 [10]

Answer:

Total Cost is 41.18 AED

Explanation:

Per day number of hours light bulbs were on=3hours

5 bulbs power consumption = 150Wx5 = 750W

Electricity usage per day = 750 x 3 = 2250Wh = 2.25kWh

April has 30 days and May has 31 days. Therefore totals days = 61days

Total power consumption = 2.25x 61 = 137.25kWh

Total Cost = 137.25x 0.30 = 41.175 AED

3 0

3 years ago

What is the SI (metric) unit of FORCE?<br><br> A. meter<br> B. newton

Tresset [83]

What is the SI (metric) unit of FORCE?

B. newton

with symbol ( N )

All the best !

4 0

3 years ago

Read 2 more answers

A toy race car travels down a 25cm ramp in 5 seconds. Calculate its speed. A. 125 cm/sec B. 5 cm/sec C. 0.2 cm/sec D. 20 cm/sec

kipiarov [429]

This would be B, 5cm/sec.
To get this answer you would need to divided 25 which is how large the ramp is, and 5 seconds, the amount of time it took to travel down the ramp.

7 0

3 years ago

Suppose that while moving into an apartment you move a refrigerator into place by sliding it across the ﬂoor. The refrigerator w

sweet-ann [11.9K]

Answer:

358 N

Explanation:

$F=\mu N$ where F is the force, $\mu$ is the coefficient of static friction between the ﬂoor and the refrigerator and N is the weight

Normally, N=mg hence $F=\mu mg$ where m is the mass of object and g is the acceleration due to gravity

In this case, N is given as 895 N and the coefﬁcient of static friction between the ﬂoor and the refrigerator is 0.400 hence substituting them in the formula we obtain

$F=\mu N= 0.4\times 895 N= 358 N$

4 0

4 years ago