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Lina20 [59]
3 years ago
9

An ideal spring of negligible mass is 11.00cm long when nothing is attached to it. When you hang a 3.05-kg weight from it, you m

easure its length to be 12.40cm .
If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.
Express your answer numerically. If there is more than one answer, enter each answer, separated by a comma.
=
Physics
1 answer:
mart [117]3 years ago
7 0

Answer

0.2067m or 0.2067m

Explanation;

Let lenght of spring= Lo= 11cm=0.110m

It is hang from a mass of

3.05-kg having a length of L1= 12.40cm= 0.124m

Force required to stretch the spring= Fkx

But weight of mass mg= kx then K= Mg/x

K= 3.05-kg× 9.8)/(0.124m-.110m)

K=2135N

But potential Energy U= 0.5Kx

X=√ 2U/k

√(2*10)/2135

X=0.0967m

The required new length= L2= L0 ±x

=

.110m ± 0.0967m

X= 0.2067m or 0.2067m hence the total lenghth

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A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
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a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is 8.75 m/s^2 in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

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W=(4)(9.8)=39.2 N

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

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c)

We said there are two forces acting in the horizontal direction:

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d)

The net force is obtained as the resultant  of the net forces in the horizontal and vertical direction. However, we have:

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Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

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\sum F = 35 N (in the forward direction)

m = 4 kg

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a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2 (forward)

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