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Mashutka [201]
2 years ago
11

A 40.0 N crate is pulled up a 5.0 m inclined plane. The angle of inclination of the plane to horizontal is 37o. If there is a co

nstant force of friction of 10.0 N between the crate and the surface, what is the net gain in potential energy by the crate? Group of answer choices 120 J -120 J 210 J -210 J
Physics
1 answer:
Digiron [165]2 years ago
4 0

The net gain in potential energy by the crate is 120 J.

<h3>What is potential energy?</h3>

Potential energy is the energy that an object has stored in it as a result of its position or condition. Potential energy exists in a stretched spring, a book held above your head, and a bicycle on top of a hill. The joule, which is represented by the letter "J," is the standard unit for calculating potential energy.

<h3 /><h3>Calculation:</h3>

We use the idea of potential energy, which is given by the equation,

PE = mgh

where,

m = mass of the crate

g = gravitational acceleration

h = height

Given,

mg = 40 N

L = 5 m

angle of inclination, ∅ = 37°

h = L sin∅

h = 5 (sin 37°)

h = 5 (3/5)

h = 3 m

Put the values in the above formula,

PE = mgh

PE = 40 (3)

PE = 120 J

Therefore, the net gain in potential energy by the crate is 120 J.

<h3 />

Learn more about the calculation of work and force here:

brainly.com/question/28043403

#SPJ4

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Answer:

a. The station is rotating at 1.496 \frac{rev}{min}

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Explanation:

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<h3>a. </h3>

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\omega  = 0.1567 \frac{rad}{s}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

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<h3>b. </h3>

The rotational speed needed is :

9.8 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.08909 \frac{rad^2}{s^2} }

\omega  = 0.2985 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.2985 \frac{rev}{min}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 2.8502 \frac{rev}{min}

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