A 178 g
 block is launched by compressing a spring of constant k=
200N/m
 a distance of 15 cm
. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction μ=0.27
. This frictional surface extends 85 cm
, followed by a frictionless curved rise, After launch, where does the block finally come to rest? Measure from the left end of the frictional zone. answer in cm

### Step 1: Calculate the initial energy stored in the spring

The potential energy stored in the compressed spring can be calculated using the formula for elastic potential energy:

\[
E_s = \frac{1}{2} k x^2
\]

Where:
- \( k = 200 \, \text{N/m} \) is the spring constant,
- \( x = 15 \, \text{cm} = 0.15 \, \text{m} \) is the compression of the spring.

Substitute the values:

\[
E_s = \frac{1}{2} \times 200 \times (0.15)^2
\]
\[
E_s = 0.5 \times 200 \times 0.0225 = 2.25 \, \text{J}
\]

### Step 2: Energy dissipated by friction

The block will travel a distance \( d = 85 \, \text{cm} = 0.85 \, \text{m} \) on the frictional surface. The work done by friction is given by:

\[
W_f = f_{\text{friction}} \times d = \mu m g \times d
\]

Where:
- \( \mu = 0.27 \) is the coefficient of friction,
- \( m = 178 \, \text{g} = 0.178 \, \text{kg} \) is the mass of the block,
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Substitute the values:

\[
W_f = 0.27 \times 0.178 \times 9.8 \times 0.85
\]
\[
W_f = 0.27 \times 0.178 \times 8.33 = 0.3996 \, \text{J}
\]

### Step 3: Determine if the block stops on the frictional surface

The total energy initially stored in the spring is 2.25 J, and the energy dissipated by friction after traveling the full 0.85 m on the frictional surface is only 0.3996 J. Since this is less than the initial energy, the block will still have energy left after crossing the frictional surface.

### Step 4: Calculate remaining energy after crossing the frictional surface

The remaining energy after friction is:

\[
E_{\text{remaining}} = E_s - W_f = 2.25 - 0.3996 = 1.8504 \, \text{J}
\]

This remaining energy is converted into kinetic energy as the block moves beyond the frictional surface.

### Step 5: Distance traveled on the frictional surface

The block will lose energy to friction as it moves on the surface, and it will stop when the kinetic energy is completely dissipated by friction. Using the remaining energy, we can calculate how much farther the block travels:

The frictional force is:

\[
f_{\text{friction}} = \mu m g = 0.27 \times 0.178 \times 9.8 = 0.4705 \, \text{N}
\]

The work done by friction is equal to the remaining kinetic energy:

\[
f_{\text{friction}} \times d_{\text{friction}} = E_{\text{remaining}}
\]

Substitute the values:

\[
0.4705 \times d_{\text{friction}} = 1.8504
\]
\[
d_{\text{friction}} = \frac{1.8504}{0.4705} = 3.93 \, \text{m}
\]

Since this distance is greater than the length of the frictional surface (0.85 m), the block will continue moving beyond the frictional zone.

### Step 6: Final resting position

After traveling the entire 0.85 m of the frictional zone, the block will have traveled an additional distance of:

\[
d_{\text{friction}} - 0.85 = 3.93 - 0.85 = 3.08 \, \text{m}
\]

Thus, the block will come to rest 393 cm from the left end of the frictional surface.

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### Summary:
The block comes to rest **393 cm** from the left end of the frictional zone.

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Would you like further clarification on any step? Here are five questions to help expand on the topic:

1. How is energy conserved in systems with friction?
2. How does the spring constant influence the energy stored in a spring?
3. What other factors could affect the stopping distance in this scenario?
4. How would the result change if the mass of the block were different?
5. How can we experimentally determine the coefficient of friction in a real-world scenario?

**Tip:** Always ensure you convert units properly, especially in problems mixing cm, m, and g, as it significantly affects calculations.

A 178 g block is launched by compressing a spring of constant k = 200 N/m a distance of 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction μ = 0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise. After launch, where does the block finally come to rest? Measure from the left end of the frictional zone. Answer in cm.

This problem involves calculating the final resting position of a 178 g block launched by a spring with k = 200 N/m, accounting for friction (μ = 0.27). Learn how to apply the work-energy principle and energy conservation with detailed formulas and steps. Suitable for Grades 10-12 physics students.

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