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2 years ago

How much work is done lifting a 9.10-kg box straight up onto a shelf that is 1.80 m high

Physics

1 answer:

love history [14]2 years ago

5 0

P= 10m = 10 x 9.10 = 91 N

Work is done lifting a 9.10-kg box straight up onto a shelf that is 1.80 m high :

A= Ph = 91 x 1.80 = 163.8 J

ok done. Thank to me :>

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An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring

ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

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<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

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<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

$F = k x$

$mg = k x$

$k = mg \div x$

$k = 1.25(9.8) \div 0.0275$

$k = 445 \frac{5}{11} \texttt{ N/m}$

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<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek$

$Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2$

$Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh$

$Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)$

$\boxed {Ek \approx 1.20 \texttt{ J}}$

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<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

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2 years ago

A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener

Elina [12.6K]

Answer:

Explanation:

From the given information:

The initial PE $(PE)_i$ = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = $P.E_f -P.E_i$

ΔP.E = 0 - $(PE)_i$

ΔP.E = $-P.E_i$

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

$\Delta P.E + \Delta K.E + \Delta U = 0$

$\Delta U = -\Delta P.E - \Delta K.E$

this can be re-written as:

$\Delta U =- (-\Delta P.E_i) - \Delta K.E$

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

$\Delta U =(70\%) \Delta P.E_i-0$

$\Delta U =(0.70) (490.5)$

ΔU = 343.35 J

Thus, the magnitude of the increase is = 343.35 J

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