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2 years ago

6

a body builder uses a chest expander with 5 strings.it takes a force of 200n to pull one string by 15cm.how much force will be n

eeded to expand the expander by 15 cm during a workout session

1 answer:

wolverine [178]2 years ago

5 0

Answer:

<u>1000N</u>

Explanation:

It takes 200N to expand 15cm of one string.

For 5 strings, force required =

200 x 5 = <u>1000N</u>

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<h3>Answer:</h3>

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1 year ago

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3 years ago

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3 years ago

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m_a_m_a [10]

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Explanation:

(I will not include units in calculations)

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Let the 2.2kg block equal $m_2$ , and the 20kg block equal $m_1$ .

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$\sum\tau=F_t_1-F_t_2=.5Ma$

This step is important: Rearrange the force summation equation to solve for each tension force.

$F_t_2=m_2a+F_f+F_g_x\\F_t_1=m_1g=m_1a$

Perform Substitution: $\sum\tau=m_1g-m_1a-(m_2a+F_f+F_g_x)=.5Ma$

Now, we need to find the friction force and the horizontal component of the force of gravity.

Note that $F_f=$ μ $F_n$

And based on our earlier summation equation: $F_n=F_g_y$

First, break $F_g$ into x and y components. $F_g_y=F_g\cos(\theta)$ , $F_g_x=F_g\sin(\theta)$

Perform substitution with this and the fact that $F_g=mg$ .

$\sum\tau=m_1g-m_1a-(m_2a+\mu*m_2g\cos(\theta)+m_2g\sin(\theta))=.5Ma$

Solving for a, plugging in numbers yields an answer of 7.23407 $\frac{m}{s^2}$

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3 years ago