
A solid iron sphere rolls down an inclined plane, while an identical hollow sphere B of same mass slides down the plane in a frictionless manner. At the bottom of the inclined plane, the total kinetic energy of sphere A is
A.) Less than that of B
B.) Equal to that of B
C.) More than that of B
D.) Sometimes more and sometimes less
Answer
600.6k+ views
Hint: The change in kinetic energy of the sphere rolling down from the inclined plane is given by the work done on the sphere. There is no frictional force acting on them because of rolling.
Formula used:
Work energy theorem for a mechanical system is given as
$W = \Delta K = {K_2} - {K_1}$
where W is the work done due to all the forces acting on the system and $\Delta K$ signifies the change in kinetic energy of the system where ${K_1}$ represents the initial kinetic energy of the system while ${K_2}$ represents the final kinetic energy of the system.
Detailed step by step solution:
When a sphere rolls down an inclined plane, there is no frictional force acting on it because of rolling.
We are given two spheres A and B. Sphere A is made of solid iron while sphere B is hollow made of iron. The two spheres are identical in size and shape. Both of them are allowed to roll down an inclined plane in a frictionless manner.
Since there is no friction acting, the only force acting is due to weight of the object and change in its height. If m is mass of sphere A and equal to the mass of sphere B while h is their height above the ground on the top of the inclined plane , then we can write the work energy theorem for two spheres as follows:
(The initial kinetic energy of the spheres zero.)
Sphere A: $mgh = {K_{{A_2}}} - 0{\text{ }} \Rightarrow {K_{{A_2}}} = mgh{\text{ }}...{\text{(i)}}$
Similarly, for sphere B: $mgh = {K_{{B_2}}} - 0{\text{ }} \Rightarrow {K_{{B_2}}} = mgh{\text{ }}...{\text{(ii)}}$
From equation (i) and (ii), we can say that the kinetic energy of two spheres is equal when they reach at the bottom of the inclined plane irrespective of the hollowness of the sphere B.
Note: Rolling is different from sliding. In sliding, an object is getting rubbed on the surface on which it is moving whereas in the case of rolling there is no rubbing taking place between the surface and the rolling object.
Formula used:
Work energy theorem for a mechanical system is given as
$W = \Delta K = {K_2} - {K_1}$
where W is the work done due to all the forces acting on the system and $\Delta K$ signifies the change in kinetic energy of the system where ${K_1}$ represents the initial kinetic energy of the system while ${K_2}$ represents the final kinetic energy of the system.
Detailed step by step solution:
When a sphere rolls down an inclined plane, there is no frictional force acting on it because of rolling.
We are given two spheres A and B. Sphere A is made of solid iron while sphere B is hollow made of iron. The two spheres are identical in size and shape. Both of them are allowed to roll down an inclined plane in a frictionless manner.
Since there is no friction acting, the only force acting is due to weight of the object and change in its height. If m is mass of sphere A and equal to the mass of sphere B while h is their height above the ground on the top of the inclined plane , then we can write the work energy theorem for two spheres as follows:
(The initial kinetic energy of the spheres zero.)
Sphere A: $mgh = {K_{{A_2}}} - 0{\text{ }} \Rightarrow {K_{{A_2}}} = mgh{\text{ }}...{\text{(i)}}$
Similarly, for sphere B: $mgh = {K_{{B_2}}} - 0{\text{ }} \Rightarrow {K_{{B_2}}} = mgh{\text{ }}...{\text{(ii)}}$
From equation (i) and (ii), we can say that the kinetic energy of two spheres is equal when they reach at the bottom of the inclined plane irrespective of the hollowness of the sphere B.
Note: Rolling is different from sliding. In sliding, an object is getting rubbed on the surface on which it is moving whereas in the case of rolling there is no rubbing taking place between the surface and the rolling object.
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