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BigorU [14]
3 years ago
15

A student pushes a box across a rough horizontal floor. If the amount of work done by the student on the box is 100 J and the am

ount of energy dissipated by friction is 40 J, what is the change in kinetic energy of the box
Physics
1 answer:
Archy [21]3 years ago
4 0

Answer:

60J

Explanation:

friction is the opposition to the motion of an object. Before a body can be set in motion, part of the energy or work-done must be used to overcome friction.

While kinetic energy is the energy associated with a body due to its motion.

Since friction tends to oppose the motion of a body,when a body start to move, some of the energy applied to the body to cause its motion is converted to heat due to friction.

Hence if a work-done of 100J is applied to a body that requires 40J of the work to overcome friction, the total change in the kinetic energy can be expressed as

K.E=100J-40J\\K.E=60J\\

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Which formula can be used to solve problems related to the first law of thermodynamics? q = w u q = u – deltaw q = u w q = u – w
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"Q = ΔU + W" is the equation is used to solve the questions related to "First law of thermodynamics".

<h3> What is the first law of thermodyanamics?</h3>

"First law of thermodynamics" states that "energy" neither created nor destroyed, but it can transfer from "one form of energy" to "another form of energy".

This "First law of thermodynamics" is also called as "law of conversation of energy". The formula for "First law of thermodynamics" of a system is that "change in internal energy of a system" is same as the difference of "heat energy" flows across the " boundaries of a system" and the "work done" on the system.

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Q = ΔU + W

Where, "ΔU" is "change in internal energy", "Q" is "heat transferred and "W" is "work done.

Hence "Q = ΔU + W" is the equation is used to solve the questions related to "First law of thermodynamics".

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brainly.com/question/15071682

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A group of students performed a compression experiment where they placed weights on top of a cylinder of material and measured t
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The material that the cylinder is made from is Butyl Rubber.

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Young's modulus, or the modulus of elasticity in tension or compression, is a mechanical property that measures the tensile or compressive strength of a solid material when a force is applied to it.

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E = \frac{stress}{strain} \\\\E = \frac{F/A}{e/l} \\\\E = \frac{Fl}{Ae} \\\\

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When 5 kg mass is applied, the extension = 10 cm - 9.61 cm = 0.39 cm = 0.0039 m.

E = \frac{(5\times 9.8) \times 0.1}{0.00126 \times 0.0039} \\\\E = 9.97 \times 10^5 \ N/m^2\\\\E = 0.000997 \times 10^9 \ N/m^2\\\\E = 0.000997 \ GPa\\\\E \approx 0.001 \ GPa

When the mass is 50 kg,

extension = 10 cm - 7.73 cm = 2.27 cm = 0.0227 m

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The Young's modulus is between 0.001 GPa  to 0.002 GPa

Thus, the material that the cylinder is made from is Butyl Rubber.

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