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A 235 \ g block is pressed against a spring of force constant 1.55 \ kN/m until the block...

Question:

A {eq}235 \ g {/eq} block is pressed against a spring of force constant {eq}1.55 \ kN/m {/eq} until the block compresses the spring {eq}10.0 \ cm {/eq}. The spring rests at the bottom of a ramp inclined at {eq}60^\circ {/eq} to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops

(a) if the ramp exerts no friction force on the block and

(b) if the coefficient of kinetic friction is {eq}0.380 {/eq}.

Energy conservation and Friction:

The energy conservation means that the total energy of the object remains constant at every instant of the system. The energy lost in the friction depends upon the friction force.

Answer and Explanation:

Given data:

  • Mass of the block {eq}\rm (m) = 0.235 \ kg {/eq}
  • Spring constant {eq}\rm (k) = 1.55 \ kN/m {/eq}
  • Compression of the spring {eq}\rm (x) = 10 \ cm {/eq}
  • Angle of the incline {eq}\rm (\theta) = 60^\circ {/eq}

(a)

The initial energy of the block

{eq}\rm E = \dfrac{1}{2}kx^{2} \\ E = \dfrac{1}{2}(1.55 \times 10^{3})(0.1)^{2} \\ E = 7.75 \ J {/eq}

Now, let us say that it has move upward by distance "d", therefore the height raised of the block would be

{eq}\rm h = d\sin\theta \\ h = d\sin60^\circ \\ h = 0.866d {/eq}

Now, applying the energy conservation

{eq}\rm E = mgh \\ 7.75 = ( 0.235) (9.8)(0.866d) \\ d = 3.88 \ m {/eq}


(b)

In this case we have to consider the friciton as well, therefore

{eq}\rm E = mgh + (\mu_{k}mg\cos\theta \times d ) \\ 7.75 = mg(0.866d + 0.38 \cos60^\circ d ) \\ d = 3.187 \ m {/eq}


Learn more about this topic:

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Energy Conservation and Energy Efficiency: Examples and Differences

from Geography 101: Human & Cultural Geography

Chapter 13 / Lesson 9
108K

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