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denis23 [38]
3 years ago
6

Lou’s latest invention, aimed at urban dog owners, is the X-R-Leash. It is made of a rubber-like material that exerts a force Fx

= (−5.7 N/m) x − (78 N/m2 ) x 2 when it is stretched a distance x. The ad claims, "You’ll never go back to your old dog leash after you’ve had the thrill of an X-R-Leash experience. And you’ll see a new look of respect in the eyes of your proud pooch." Find the work done on a dog by the leash if the person remains stationary, and the dog bounds off, stretching the X-R-Leash from x = 0 m to x = 22 m. Answer in units of kJ.
Physics
1 answer:
GaryK [48]3 years ago
5 0

Answer:

W = 29.06 KJ

Explanation:

The work done while stretching the leash can be calculated by the following formula:

W = \int\limits^b_a {F_{x}} \, dx \\

whee,

W = Work Done = ?

Fₓ = Forcing Function = (-5.7 N/m)x - (78 N/m²)x²

a = starting point of x = 0 m

b = end point of x = 22 m

Therefore,

W = \int\limits^{22\ m}_{0\ m} {(-(5.7\ N/m)x - (7.8\ N/m^{2})x^{2}}) \, dx \\Integrating\ we\ get:\\W = -\frac{(5.7\ N/m)x^{2}}{2} - \frac{(7.8\ N/m^{2})x^{3}}{3}\\Applying\ limits:\\W = -\frac{(5.7\ N/m)(22\ m)^{2}}{2} - \frac{(7.8\ N/m^{2})(22\ m)^{3}}{3} - 0\\W = - 1379.4\ J - 27684.8\ J\\W = 29064.2\ J

<u>W = 29.06 KJ</u>

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A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.
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B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

\sum F_y=m_1*a_m_i_n = T-m_1*g

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\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g

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a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: \sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a

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The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: \sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a

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T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N

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