Question

2 blocks are released from rest at the same time from a vertical height h above the ground level. Block one has mass M and falls straight down. Whereas block 2 has mass 2M and slides down a frictionless incline. Assume that air resistance is negligible.

A) Which block reaches the ground level first ?

B) when the blocks reach the ground level, which one will have a greater kinetic energy ?

Answer 1

Answer on Question #68536 Physics / Other

2 blocks are released from rest at the same time from a vertical height $h$ above the ground level. Block one has mass $M$ and falls straight down. Whereas block 2 has mass $2M$ and slides down a frictionless incline. Assume that air resistance is negligible.

A) Which block reaches the ground level first?

B) when the blocks reach the ground level, which one will have a greater kinetic energy?

Solution:

A)

Let the $t_1$ and $t_1$ are falling time for the block 1 and block 2 respectively. Thus

$$h = \frac {g t _ {1} ^ {2}}{2}, \qquad t _ {1} = \sqrt {\frac {2 h}{g}}.$$ $$l = \frac {a t _ {2} ^ {2}}{2}, \qquad t _ {2} = \sqrt {\frac {2 l}{a}}.$$ $$a = g \sin \alpha , \qquad l = \frac {h}{\sin \alpha}$$ $$t _ {2} = \sqrt {\frac {2 l}{g \sin \alpha}}, \qquad t _ {2} = \sqrt {\frac {2 h}{g \sin^ {2} \alpha}} = \frac {t _ {1}}{\sin \alpha}.$$

Because $0 < \sin \alpha < 1$ , thus $t_2 > t_1$

B) From the conservation energy law

$$K _ {1} = M g h$$

and

$$K _ {2} = 2 M g h$$

So

$$K _ {2} > K _ {1}.$$

Answer: A) the block 1 reaches the ground level first. B) the block 2 will have a greater kinetic energy.

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