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Math  /  Algebra

QuestionA 0.750 kg object vibrates at the end of a horizontal spring ( k =995 N/m=995 \mathrm{~N} / \mathrm{m} ) along a frictionless surface. The speed of the object is 2.50 m/s2.50 \mathrm{~m} / \mathrm{s} when its displacement is 0.145 m . What is the maximum displacement of the object?

Studdy Solution

STEP 1

1. The object has a mass of 0.750 kg.
2. The spring constant k k is 995 N/m.
3. The surface is frictionless.
4. The speed of the object is 2.50 m/s when its displacement is 0.145 m.
5. We are trying to find the maximum displacement of the object.

STEP 2

1. Use the conservation of mechanical energy principle.
2. Calculate the total mechanical energy at the given displacement.
3. Express the total mechanical energy in terms of maximum displacement.
4. Solve for the maximum displacement.

STEP 3

Use the conservation of mechanical energy principle. The total mechanical energy in a spring-mass system is the sum of kinetic energy and potential energy, and it remains constant.

STEP 4

Calculate the total mechanical energy at the given displacement. The kinetic energy (KE KE ) is given by:
KE=12mv2 KE = \frac{1}{2} m v^2
where m=0.750 kg m = 0.750 \, \text{kg} and v=2.50 m/s v = 2.50 \, \text{m/s} .
KE=12×0.750×(2.50)2 KE = \frac{1}{2} \times 0.750 \times (2.50)^2
KE=12×0.750×6.25 KE = \frac{1}{2} \times 0.750 \times 6.25
KE=2.34375 J KE = 2.34375 \, \text{J}
The potential energy (PE PE ) stored in the spring is given by:
PE=12kx2 PE = \frac{1}{2} k x^2
where k=995 N/m k = 995 \, \text{N/m} and x=0.145 m x = 0.145 \, \text{m} .
PE=12×995×(0.145)2 PE = \frac{1}{2} \times 995 \times (0.145)^2
PE=12×995×0.021025 PE = \frac{1}{2} \times 995 \times 0.021025
PE=10.4586875 J PE = 10.4586875 \, \text{J}
The total mechanical energy (E E ) is:
E=KE+PE E = KE + PE
E=2.34375+10.4586875 E = 2.34375 + 10.4586875
E=12.8024375 J E = 12.8024375 \, \text{J}

STEP 5

Express the total mechanical energy in terms of maximum displacement. At maximum displacement, the speed is zero, so all the energy is potential energy:
E=12kA2 E = \frac{1}{2} k A^2
where A A is the maximum displacement.

STEP 6

Solve for the maximum displacement A A .
12.8024375=12×995×A2 12.8024375 = \frac{1}{2} \times 995 \times A^2
A2=12.8024375×2995 A^2 = \frac{12.8024375 \times 2}{995}
A2=25.604875995 A^2 = \frac{25.604875}{995}
A2=0.025735 A^2 = 0.025735
A=0.025735 A = \sqrt{0.025735}
A≈0.1604 m A \approx 0.1604 \, \text{m}
The maximum displacement of the object is:
0.1604 m \boxed{0.1604 \, \text{m}}

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