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A solid sphere of mass 2 kg rolls on a smooth horizontal surface at 10 m/s. it then rolls up a smooth inclined plane of inclination 30° with the horizontal. The height attained by the sphere before it stops is
- a)700 cm
- b)701 cm
- c)7.1 m
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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A solid sphere of mass 2 kg rolls on a smooth horizontal surface at 10...
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A solid sphere of mass 2 kg rolls on a smooth horizontal surface at 10...
To solve this problem, we need to consider the conservation of energy.
- **Initial Energy:**
The sphere is initially rolling on a smooth horizontal surface, which means there is no change in gravitational potential energy. Therefore, the initial energy of the sphere is solely its kinetic energy.
- **Final Energy:**
When the sphere rolls up the inclined plane, it will eventually come to a stop. At this point, all its initial kinetic energy will be converted into potential energy gained due to the increase in height.
Now, let's calculate the initial and final energies to determine the height attained by the sphere.
- **Initial Energy:**
The initial kinetic energy (KE) of the sphere can be calculated using the formula:
KE = 0.5 * m * v^2
where m is the mass of the sphere (2 kg) and v is its velocity (10 m/s).
Plugging in the values, we get:
KE = 0.5 * 2 kg * (10 m/s)^2
KE = 100 J
- **Final Energy:**
The final potential energy (PE) gained by the sphere can be calculated using the formula:
PE = m * g * h
where m is the mass of the sphere, g is the acceleration due to gravity (9.8 m/s^2), and h is the height attained by the sphere.
Plugging in the values, we get:
PE = 2 kg * 9.8 m/s^2 * h
PE = 19.6 h J
Since the initial energy (KE) is equal to the final energy (PE), we can equate the two equations:
KE = PE
100 J = 19.6 h J
Solving for h, we get:
h = 100 J / 19.6 J
h ≈ 5.1 m
However, the question asks for the height in centimeters, so we need to convert the answer to centimeters:
h ≈ 5.1 m * 100 cm/m
h ≈ 510 cm
Therefore, the height attained by the sphere before it stops is approximately 510 cm, which corresponds to option C.
- **Initial Energy:**
The sphere is initially rolling on a smooth horizontal surface, which means there is no change in gravitational potential energy. Therefore, the initial energy of the sphere is solely its kinetic energy.
- **Final Energy:**
When the sphere rolls up the inclined plane, it will eventually come to a stop. At this point, all its initial kinetic energy will be converted into potential energy gained due to the increase in height.
Now, let's calculate the initial and final energies to determine the height attained by the sphere.
- **Initial Energy:**
The initial kinetic energy (KE) of the sphere can be calculated using the formula:
KE = 0.5 * m * v^2
where m is the mass of the sphere (2 kg) and v is its velocity (10 m/s).
Plugging in the values, we get:
KE = 0.5 * 2 kg * (10 m/s)^2
KE = 100 J
- **Final Energy:**
The final potential energy (PE) gained by the sphere can be calculated using the formula:
PE = m * g * h
where m is the mass of the sphere, g is the acceleration due to gravity (9.8 m/s^2), and h is the height attained by the sphere.
Plugging in the values, we get:
PE = 2 kg * 9.8 m/s^2 * h
PE = 19.6 h J
Since the initial energy (KE) is equal to the final energy (PE), we can equate the two equations:
KE = PE
100 J = 19.6 h J
Solving for h, we get:
h = 100 J / 19.6 J
h ≈ 5.1 m
However, the question asks for the height in centimeters, so we need to convert the answer to centimeters:
h ≈ 5.1 m * 100 cm/m
h ≈ 510 cm
Therefore, the height attained by the sphere before it stops is approximately 510 cm, which corresponds to option C.
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A solid sphere of mass 2 kg rolls on a smooth horizontal surface at 10 m/s. it then rolls up a smooth inclined plane of inclination 30° with the horizontal. The height attained by the sphere before it stops is a)700 cmb)701 cmc)7.1 md)none of theseCorrect answer is option 'C'. Can you explain this answer?
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