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3 years ago

The potential energy of a 25 kg bicycle resting at the top of a 3 m high hill is _______ J

Physics

2 answers:

HACTEHA [7]3 years ago

7 0

Hello!

The potential energy of a 25 kg bicycle resting at the top of a 3 m high hill is _______ J

If: 1 Joule = 1 kg.m²/s²

We have the following data:

GPE (Gravitational Potential Energy) = ? (in J)

m (mass) = 25 kg

g (gravity) ≈ 9.81 m/s²

h (height) = 3 m

We apply the data to the formula of the gravitational potential energy, we have:

$GPE = m*g*h$

$GPE = 25\:kg*9.81\:\dfrac{m}{s^2} *3m$

$GPE = 735.75\:kg*\dfrac{m^2}{s^2}$

$\boxed{\boxed{GPE = 735.75\:J}}\:\:\:\:\:\:\bf\green{\checkmark}$

<u><em>Answer:</em></u>

________________________

$\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

frozen [14]3 years ago

5 0

The potential energy:
E p = m · g · h
m = 25 kg, g = 9.81 m/s², h = 3 m
E p = 25 kg · 9.81 m/s² · 3 m = 735.75 J

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Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

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Answer:

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Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

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Coefficient of kinetic friction, $\mu = 0.2$

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The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

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$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

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Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

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