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4 years ago

12

A piano mover raises a 1000-n piano at a constant speed using a very light rope in a frictionless pulley system, as shown in the

figure. with what force is the mover pulling down on the rope?

2 answers:

Karolina [17]4 years ago

6 0

The magnitude of the mover's pulling force is 500 N

$\texttt{ }$

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

weight of piano = w = 1000 N

acceleration of piano = a = 0 m/s² → <em>constant speed</em>

<u>Asked:</u>

magnitude of the pulling force = T = ?

<u>Solution:</u>

<em>Firstly , we will draw </em><em>the free body diagram</em><em> as shown in the attachment.</em>

<em>Next , we will calculate the pulling force by using </em><em>Newton's Law of Motion</em><em> as follows:</em>

$\Sigma F = ma$

$T + T - w = ma$

$2T - w = ma$

$2T = w + ma$

$T = ( w + ma ) \div 2$

$T = ( 1000 + m(0) ) \div 2$

$T = 1000 \div 2$

$T = 500 \texttt{ N}$

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Newton's Law of Motion: brainly.com/question/10431582
Example of Newton's Law: brainly.com/question/498822

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Rina8888 [55]4 years ago

5 0

Here As we can see the figure that the end of the rope is pulled by some force F

Now as we can see that Piano is connected by a pulley which is passing over the pulley so effectively net force on the piano upwards will be 2F as it is connected by 2 ropes by the pulley

Now for constant velocity of the piano we will say

$2F - mg = ma$

since velocity is constant so acceleration must be ZERO

so here we have

$2F - mg = 0$

as we know here that

mg = 1000 N

so we will have

$2F - 1000 = 0$

$F = 500 N$

so here force must be 500 N

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Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

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∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

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$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

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$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

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To verify the units:

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↓ ↓ ↓

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Thus, the units are verified.

The integrated charge Q

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$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

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