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A 205-g block is pressed against a spring of force constant 1.18 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0° to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops under the following conditions

User Prid
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1 Answer

4 votes

Answer:

L = 3.391m

Step-by-step explanation:

a) Given:

Mass of the block, m = 205g=0.205kg

spring constant, k = 1.18 kN/m

Angle made by the ramp = 60°

displacement of the spring, x = 10cm= 0.1m

Now,

The initial potential energy of the spring = final gravitational potential energy of block

mathematically,


(1)/(2)kx^2 = mgh

where,

g = acceleration due to gravity

h = height of the block above the ground = LsinΘ

L is the displacement of the block parallel to the ramp


(1)/(2)kx^2 = mgLsin\theta

substituting the values in the above equation we get


(1)/(2)1.18* 1000* 0.1^2 = 0.205* 9.8* Lsin60^o

or


5.9 = 1.739* L

or

L = 3.391m

User Sadidul Islam
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