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Art [367]
3 years ago
10

A box slides down a frictionless ramp.if it starts at rest, what is it’s speed at the bottom?

Physics
1 answer:
zhenek [66]3 years ago
3 0

8.854 m/s is the speed of the box after it reaches bottom of the ramp.

<u>Explanation</u>:

From the figure we came to know that height of the block is 4 m.

We know that,

Total "initial energy of an object" = Total "final energy of an object "

Total "initial energy of an object" is = "sum of potential energy" and "kinetic energy" of an object at its initial position.

\text { "g" acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}

\text { Total initial energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{i}}+\frac{1}{2} \mathrm{m} v_{i}^{2}

Initial velocity is “0” as the object does not have starting speed

\text { Height of the block where the object is placed initially }\left(h_{i}\right) \text { is } 4 \mathrm{m} \text { . }

\text { Total initial energy }=\mathrm{m} \times 9.8 \times 4+\frac{1}{2} \mathrm{m} 0^{2}

Total initial energy = 39.2 × m

\text { Total final energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{f}}+\frac{1}{2} m v_{f}^{2}

\text { We need to find final velocity } v_f

\text { Height of the block where the object is travelled to bottom (h_) is } 0 \mathrm{m} \text { . }

\text { Total final energy }=\mathrm{m} \times 9.8 \times 0+\frac{1}{2} m v_{f}^{2}

Now,  Total initial energy of an object = Total final energy of an object

39.2 \times \mathrm{m}=0.5 \mathrm{m} v_{f}^{2}

\frac{39.2}{0.5}=v_{f}^{2}

v_{f}^{2}=78.4

v_{f}=\sqrt{78.4}

v_{f}=8.854 \mathrm{m} / \mathrm{s}

Final speed is 8.854 m/s.

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Which of the following would most likely cause a decrease in the quality supplied
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2 years ago
Identify evidence that supports the theories of continental drift and plate tectonics. put responses in the correct input to ans
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6 0
2 years ago
A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th
ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = \frac{1}{2}mv^2.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 \frac{1}{2}mv^2=\ mgh

                ⇒ v^2=\ 2gh

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                ⇒ h=\ 29.88\ m

Hence, the height of the cliff is 29.88 m

             


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3 years ago
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katrin [286]
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