Mechanics - 6

Classical Mechanics Level 2

A block starts with a velocity of 5 m/s \text{5 m/s} at A A , which is at a height 10 m \text{10 m} , ends just reaching B B , which is at a height h m h \text{ m} . If all the surfaces are frictionless, find h h . Take g = 10 m/s 2 g = \text{10 m/s}^2 .

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The answer is 11.25.

Ram Mohith by Ram Mohith , 18, India

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2 solutions

Chew-Seong Cheong
Jan 26, 2019

Neglect all friction and air resistance, by conservation of energy , the energy of the block at point B B equals to that at the initial position of A A , that is:

E B = E A m g h + 1 2 m v B 2 = m g h A + 1 2 m v A 2 where h A is the height at the block at point A g h + 0 = 10 g + 5 2 2 and v A , v B are the respective velocities at A and B . ⟹ h = 10 + 25 20 = 11.25 \begin{aligned} E_B & = E_A \\ mgh + \frac 12 mv_B^2 & = mgh_A + \frac 12 mv_A^2 & \small \color{#3D99F6} \text{where }h_A \text{ is the height at the block at point A} \\ gh + 0 & = 10g + \frac {5^2}2 & \small \color{#3D99F6} \text{and }v_A, v_B \text{ are the respective velocities at }A \text{ and }B. \\ \implies h & = 10 + \frac {25}{20} \\ & = \boxed{11.25} \end{aligned}

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Miss Physicist
Feb 16, 2019

We know that total initial energy is equal to total final energy.
Let the mass of object be x.
Then we can say that 100x+12.5x=0+10xh. (“Just reaching” is highlighted to show that at the final position the body is at rest and hence kinetic energy is zero).
112.5x=10xh


HENCE THE ANSWER IS 11.25 m

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