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nirvana33 [79]
3 years ago
13

A rock has a mass of 25kg. The acceleration due to gravity is 10m/s and the height of the rock is 42 m. Find the potential energ

y of the rock
Physics
2 answers:
alisha [4.7K]3 years ago
7 0

Answer:

 The potential energy of the rock = 10.5 kN

Explanation:

 Mass of rock = 25 kg

 Acceleration due to gravity = 10 m/s²

 Height = 42 m

 Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity and h is the height.

 PE = 25 x 10 x 42 = 10500 N = 10.5 kN

 The potential energy of the rock = 10.5 kN

irina [24]3 years ago
7 0

Answer:

The answer is Potential energy is equal to 10500 Joules

Step by step Explanation:

1) Potential Energy is the energy that a body located at a certain height above the ground has.

2) The potential energy is expressed by the formula PE = mgh where (PE) is the potential energy (m) represents the mass, (g) represents the acceleration of gravity, and (h) represents the height.

3) Knowing this we can continue to replace the values given in the problem:

PE = (25Kg)(10m/s^2)(42m)

PE = 10500J

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Kinetic Energy = 1/2*mv²,  Where m is mass in kg, v is velocity in m/s

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1/2*mv² = E

1/2*m*30² = 33750

m = (2*33750) / (30²)     Using a calculator

m = 75 kg

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A solid, uniform disk of radius 0.250 m and mass 45.2 kg rolls down a ramp of length 5.40 m that makes an angle of 17.0° with th
serious [3.7K]

Explanation:

Given that,

Radius of the disk, r = 0.25 m

Mass, m = 45.2 kg

Length of the ramp, l = 5.4 m

Angle made by the ramp with horizontal, \theta=17^{\circ}

Solution,

As the disk starts from rest from the top of the ramp, the potential energy is equal to the sum of translational kinetic energy and the rotational kinetic energy or by using the law of conservation of energy as :

(a) mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

h is the height of the ramp

sin\theta=\dfrac{h}{5.4}

h=sin(17)\times 5.4=1.57\ m

v is the speed of the disk's center

I is the moment of inertia of the disk,

I=\dfrac{1}{2}mr^2

\omega=\dfrac{v}{r}

mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}\dfrac{1}{2}mr^2\times (\dfrac{v}{r})^2

gh=\dfrac{1}{2}v^2+\dfrac{1}{4}v^2

gh=\dfrac{3}{4}v^2

9.8\times 1.57=\dfrac{3}{4}v^2

v = 4.52 m/s

(b) At the bottom of the ramp, the angular speed of the disk is given by :

\omega=\dfrac{v}{r}

\omega=\dfrac{4.52}{0.25}

\omega=18.08\ rad/s

Hence, this is the required solution.

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3 years ago
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