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Q-4: A block of mass

m

is released fr...

Apr 8, 2024

Download solution Problem

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Q-4: A block of mass

m

is released from rest and falls a distance

h

before contacting the spring on the left, as shown, and then falls further down a distance of

100 \mathrm{~mm}

before contacting a second spring on the right, as shown. The stiffness of the spring on the left is

1000 \mathrm{~N} / \mathrm{m}

, and the stiffness of the spring on the right is

2000 \mathrm{~N} / \mathrm{m}

. If the maximum compression of the spring on the left is

200 \mathrm{~mm}

, determine the value of h. The mass

m=9 \mathrm{~kg}

Solution

Calculate the potential energy at height

h

PE = mgh

Calculate the elastic potential energy at maximum compression for the left spring:

PE_{\text{spring}} = \frac{1}{2}kx^2

where

k = 1000 \text{ N/m}

and

x = 0.2 \text{ m}

Set the potential energy equal to the elastic potential energy to find

h

mgh = \frac{1}{2}kx^2

Solve for

h

h = \frac{\frac{1}{2}kx^2}{mg}

Substitute the given values:

h = \frac{\frac{1}{2}(1000 \text{ N/m})(0.2 \text{ m})^2}{(9 \text{ kg})(9.81 \text{ m/s}^2)}

Calculate

h

h = \frac{\frac{1}{2}(1000)(0.04)}{(9)(9.81)}

Simplify to find the value of

h

h = \frac{20}{88.29} \approx 0.226 \text{ m}

Answer

h \approx 0.226 \text{ m}

Key Concept

Conservation of Mechanical Energy: In the absence of non-conservative forces, mechanical energy is conserved. This means that the potential energy at the height

h

will be converted into the elastic potential energy of the spring when the block is at maximum compression.

Explanation

The block's potential energy due to its height is transformed into elastic potential energy of the spring upon compression. By equating the two energies, we can solve for the initial height

h