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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)70 cm
- b)701 cm
- c)7 m
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A solid sphere of mass 2 kg rolls on a smooth horizontal surface at 10...
Using the energy conservation
we know ωr = v = 10
we know ωr = v = 10
Most Upvoted Answer
A solid sphere of mass 2 kg rolls on a smooth horizontal surface at 10...
We can use the principle of conservation of mechanical energy to solve this problem.
Initially, the sphere is rolling on a smooth horizontal surface. Since there is no friction, the only energy the sphere has is its kinetic energy.
The kinetic energy (KE) of a rolling sphere is given by:
KE = 1/2 * I * ω^2
where I is the moment of inertia and ω is the angular velocity.
For a solid sphere, the moment of inertia is given by:
I = 2/5 * m * r^2
where m is the mass of the sphere and r is the radius.
Given that the mass of the sphere is 2 kg, we can calculate the moment of inertia:
I = 2/5 * 2 kg * r^2
I = 8/5 * r^2 kg.m^2
Next, we need to find the angular velocity ω. The angular velocity of a rolling sphere is related to its linear velocity v by the equation:
v = ω * r
where v is the linear velocity and r is the radius.
Given that the linear velocity is 10 m/s, we can calculate the angular velocity:
10 m/s = ω * r
Now, the sphere rolls up a smooth inclined plane of inclination 30 degrees. Since there is no friction, the only energy the sphere has is its potential energy.
The potential energy (PE) of the sphere at the top of the inclined plane is given by:
PE = m * g * h
where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height of the inclined plane.
Given that the mass of the sphere is 2 kg, the acceleration due to gravity is 9.8 m/s^2, and the height of the inclined plane can be calculated using the equation:
h = sin(30) * d
where d is the length of the inclined plane.
Finally, we can set the initial kinetic energy of the sphere equal to its potential energy at the top of the inclined plane to find the height:
1/2 * I * ω^2 = m * g * h
Substituting the expressions for I, ω, and h, we can solve for the height:
1/2 * (8/5 * r^2) * (10 m/s / r)^2 = 2 kg * 9.8 m/s^2 * sin(30) * d
Simplifying and solving for d:
(8/5 * r^2) * (10^2 m^2/s^2 / r^2) = 2 kg * 9.8 m/s^2 * 0.5 * d
8 * 10^2 / 5 = 2 * 9.8 * 0.5 * d
160 = 9.8 * d
d = 160 / 9.8
d ≈ 16.33 meters
Therefore, the height of the inclined plane is approximately 16.33 meters.
Initially, the sphere is rolling on a smooth horizontal surface. Since there is no friction, the only energy the sphere has is its kinetic energy.
The kinetic energy (KE) of a rolling sphere is given by:
KE = 1/2 * I * ω^2
where I is the moment of inertia and ω is the angular velocity.
For a solid sphere, the moment of inertia is given by:
I = 2/5 * m * r^2
where m is the mass of the sphere and r is the radius.
Given that the mass of the sphere is 2 kg, we can calculate the moment of inertia:
I = 2/5 * 2 kg * r^2
I = 8/5 * r^2 kg.m^2
Next, we need to find the angular velocity ω. The angular velocity of a rolling sphere is related to its linear velocity v by the equation:
v = ω * r
where v is the linear velocity and r is the radius.
Given that the linear velocity is 10 m/s, we can calculate the angular velocity:
10 m/s = ω * r
Now, the sphere rolls up a smooth inclined plane of inclination 30 degrees. Since there is no friction, the only energy the sphere has is its potential energy.
The potential energy (PE) of the sphere at the top of the inclined plane is given by:
PE = m * g * h
where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height of the inclined plane.
Given that the mass of the sphere is 2 kg, the acceleration due to gravity is 9.8 m/s^2, and the height of the inclined plane can be calculated using the equation:
h = sin(30) * d
where d is the length of the inclined plane.
Finally, we can set the initial kinetic energy of the sphere equal to its potential energy at the top of the inclined plane to find the height:
1/2 * I * ω^2 = m * g * h
Substituting the expressions for I, ω, and h, we can solve for the height:
1/2 * (8/5 * r^2) * (10 m/s / r)^2 = 2 kg * 9.8 m/s^2 * sin(30) * d
Simplifying and solving for d:
(8/5 * r^2) * (10^2 m^2/s^2 / r^2) = 2 kg * 9.8 m/s^2 * 0.5 * d
8 * 10^2 / 5 = 2 * 9.8 * 0.5 * d
160 = 9.8 * d
d = 160 / 9.8
d ≈ 16.33 meters
Therefore, the height of the inclined plane is approximately 16.33 meters.
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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)70cmb)701 cmc)7md)none of theseCorrect answer is option 'C'. Can you explain this answer?
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