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A horizontal spring with spring constant 110 N/m is compressed 16 cm and used to launch a 2.7 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Use work and energy to find how far the box slidesacross the rough surface before stopping.

User Kapso
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2 Answers

2 votes

Final answer:

To find how far the box slides across the rough surface before stopping, we can use work and energy. By calculating the change in potential energy of the box-spring system, we can determine the work done by friction, which is equal to the force of friction times the distance traveled on the rough surface. Using the given values, the box slides approximately 0.354 meters on the rough surface before stopping.

Step-by-step explanation:

To find how far the box slides across the rough surface before stopping, we need to calculate the work done by friction during its motion. The work done by friction is equal to the change in mechanical energy of the box. Initially, the box has some energy which is converted into the potential energy of the compressed spring. When the box slides on the rough surface, some of its mechanical energy is lost to friction, resulting in a smaller compression of the spring. By using the principle of conservation of mechanical energy, we can find the distance the box slides on the rough surface before stopping.

Given:

  • Spring constant (k) = 110 N/m
  • Compression of the spring (x) = 16 cm = 0.16 m
  • Mass of the box (m) = 2.7 kg
  • Coefficient of kinetic friction (μk) = 0.15

First, let's calculate the initial potential energy of the box-spring system:

Initial potential energy (Uinitial) = 0.5 * k * x²

Plugging in the values:

Uinitial = 0.5 * 110 N/m * (0.16 m)²

Uinitial ≈ 0.5 * 110 N/m * 0.0256 m²

Uinitial ≈ 0.5 * 2.816 N*m

Uinitial ≈ 1.408 N*m

Next, let's calculate the final potential energy of the box-spring system:

Final potential energy (Ufinal) = 0.5 * k * x'², where x' is the new compression of the spring

The work done by friction is equal to the change in potential energy:

Work done by friction (Wfriction) = Uinitial - Ufinal

As the box stops moving on the rough surface, its final potential energy is zero, meaning all of its initial potential energy has been lost.

Wfriction = Uinitial - Ufinal

Wfriction = 1.408 N*m - 0 N*m

Wfriction = 1.408 N*m

The work done by friction is equal to the force of friction times the distance traveled on the rough surface:

Wfriction = frictional force * distance

Since the force of friction is equal to the normal force times the coefficient of kinetic friction:

Wfriction = (normal force) * (coefficient of kinetic friction) * distance

Now, let's find the normal force:

The normal force is equal to the weight of the box, which is given by the mass times the acceleration due to gravity:

Weight (W) = mass * gravitational acceleration

W = 2.7 kg * 9.8 m/s²

W ≈ 26.46 N

Therefore, the normal force is also 26.46 N.

Now, let's substitute the values into the equation:

1.408 N*m = 26.46 N * (0.15) * distance

1.408 N*m = 3.969 N * distance

distance = 1.408 N*m / 3.969 N ≈ 0.354 m

Therefore, the box slides approximately 0.354 meters on the rough surface before stopping.

User McShaman
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5.6k points
3 votes

Answer:

0.35 m

Explanation:

We are given that

Spring constant,
k=110N/m

Length,l=16 cm=
16* 10^(-2) m

1 m=100 cm

Mass,m=2.7 kg

Coefficient of kinetic friction,
\mu_k=0.15

Initial potential energy,
U_1=0

Final potential energy,
K_2=0

According to law of conservation of energy


K_1+U_1=K_2+U_2


(1)/(2)kx^2=\mu mg d


d=(kx^2)/(\mu mg)=(110* (16* 10^(-2))^2)/(2* 0.15* 2.7* 9.8)


d=0.35 m

User Sooobus
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5.3k points