Question

An object of mass m is released from the top of a smooth inclined plane of height h. Its speed at the bottom of the plane is proportional to :

• A.
• B. m
• C. $\mathrm { m } ^ { 2 }$
• D.

Answer 1

Answer: (A)

Answer 2

Let v is the speed of the object at the bottom of the plane.

According to work-energy theorem

$\mathrm { w } = \Delta \mathrm { K } = \mathrm { K } _ { \mathrm { f } } - \mathrm { K } _ { \mathrm { i } }$

$\mathrm { mgh } = \frac { 1 } { 2 } \mathrm { mv } ^ { 2 } - \frac { 1 } { 2 } \mathrm { mu } ^ { 2 }$ $( \therefore u = 0 )$

$\mathrm { mgh } = \frac { 1 } { 2 } \mathrm { mv } ^ { 2 }$ $( \because u = 0 )$

or $v = \sqrt { 2 g h }$

From the above expression, it is clear that v is independent of the mass of an object.