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4 years ago

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A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put

on the launcher, the launcher spring compresses by 10.0 cm. The watermelon is then pushed down by an additional 30.0 cm and it’s ready to go. Just before the launch, how much energy is stored in the spring?

1 answer:

olga_2 [115]4 years ago

6 0

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

$F_s = F_g$

$kx = mg$

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

$k = \frac{mg}{x}$

$k = \frac{8*9.8}{0.1}$

$k = 784N/m$

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

$PE = \frac{1}{2} kx^2$

$PE =\frac{1}{2} 784*0.4^2$

$PE = 63.72J$

Therefore the energy stored in the spring is 63.72J

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The battery capacity of a lithium ion battery in a digital music player is 750 mA-h. The manufacturer claims that the player can

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Answer:

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(D) is correct option.

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4 years ago

Read 2 more answers

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3 years ago

Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to

stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

$\mu_{sB}=0.126$

$\mu_{sC}=0.168$

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

$\sum \tau_{A}=0$

$N(3m)-W(1.5m)=0$

When solving for N we get:

$N=\frac{W(1.5m)}{3m}$

$N=\frac{(1962N)(1.5m)}{3m}$

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Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

$\sum F_{y}=0$

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$F_{By}=N_{c}$

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$P_{x}=360N(\frac{4}{5})=288N$

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$\sum \tau_{C}=0$

$f_{sB}(1.75m)-P_{x}(0.75m)=0$

$f_{sB}=\frac{288N(0.75m)}{1.75m}$

$f_{sB}=123.43N$

With the static friction force in point B we can find the coefficient of static friction in B:

$\mu_{sB}=\frac{f_{sB}}{N}$

$\mu_{sB}=\frac{123.43N}{981N}$

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$f_{sC}=P_{x}-f_{xB}$

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$f_{sC}=N_{c}\mu_{sC}$

and now we can use this to find static friction coefficient in point C.

$\mu_{sC}=\frac{f_{sC}}{N}$

$\mu_{sC}=\frac{164.57N}{981N}$

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