Problem 43 - Physics for Scientists and Engineers with Modern Physics

Question

(III) \(\mathrm{A} 2.0\) -kg block slides along a horizontal surface with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) The block has a speed \(v=1.3 \mathrm{m} / \mathrm{s}\) when it strikes a massless spring head- on. (a) If the spring has force constant \(k=120 \mathrm{N} / \mathrm{m},\) how far is the spring compressed? (b) What minimum value of the coefficient of static friction, \(\mu_{\mathrm{S}},\) will assure that the spring remains compressed at the maximum compressed position? (c) If \(\mu_{\mathrm{s}}\) is less than this, what is the speed of the block when it detaches from the decompressing spring? [Hint: Detach- ment occurs when the spring reaches its natural length \((x=0) :\) explain why 1

Step-by-Step Solution

Answer

(a) Use conservation of energy, solve for compression x. (b) Use static friction balance equation and solve for \( \mu_s \). (c) Use energy conservation to find block speed on detachment.

Step 1: Determine the work done by friction

The force of friction is given by \( F_f = \mu_k \cdot m \cdot g \), where \( m = 2.0 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Thus, \( F_f = 0.30 \times 2.0 \times 9.8 = 5.88 \text{ N} \). This is the force opposing the block's motion.

Step 2: Calculate the initial kinetic energy of the block

The initial kinetic energy is given by the formula \( KE = \frac{1}{2} m v^2 \). Substituting the given values, we have \( KE = \frac{1}{2} \times 2.0 \times (1.3)^2 = 1.69 \text{ J} \).

Step 3: Set up the energy conservation equation

When the block compresses the spring completely, its kinetic energy will have been converted into spring potential energy and work done against friction. The equation is: \( KE = \frac{1}{2} k x^2 + F_f x \) Where \( x \) is the compression of the spring, \( k = 120 \text{ N/m} \), and \( F_f = 5.88 \text{ N} \).

Step 4: Solve for spring compression distance

Substitute known values into the equation from Step 3: \( 1.69 = \frac{1}{2} \times 120 \times x^2 + 5.88 \times x \) This is a quadratic equation in the form \( ax^2 + bx + c = 0 \). Solve it to find the compression distance \( x \).

Step 5: Find the minimum static friction coefficient

At maximum compression, the spring force \( F_s = kx \) must equal the maximum static friction force \( F_f = \mu_s mg \). Solve \( kx = \mu_s mg \) for \( \mu_s \) by substituting the previously found \( x \), force constant \( k = 120 \text{ N/m} \), and \( m = 2.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \).

Step 6: Determine speed at detachment if \( \mu_s < \mu_s (min) \)

If \( \mu_s < \mu_s (min) \), the block will not stay compressed when the spring is at its maximum compression. When the spring decompresses and detaches the block, all potential energy converts back to kinetic energy: \( PE_{spring} = KE_{block} \). Use \( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \), solve for the velocity \( v \) at detachment.

Key Concepts

Kinetic Friction

Kinetic friction is the force that opposes the movement of two surfaces sliding past each other. It acts in the direction opposite to the motion. In this physics problem involving a 2.0 kg block, the coefficient of kinetic friction is given as \( \mu_k = 0.30 \). This coefficient helps us find the frictional force exerted on the block.

To calculate the force of kinetic friction \( F_f \), use the formula:

• \( F_f = \mu_k \times m \times g \)
• Here, \( m = 2.0 \) kg is the mass of the block, and \( g = 9.8 \) m/s² is the acceleration due to gravity.
• Substituting these values, we get \( F_f = 0.30 \times 2.0 \times 9.8 = 5.88 \) N.

This force plays a crucial role as it reduces the block's kinetic energy when moving over the surface, leading to the compression of the spring upon impact.

Spring Compression

Spring compression involves transforming the kinetic energy of the block into potential energy stored in the spring. When the block, moving at a speed of 1.3 m/s, hits the spring, it compresses it as it pushes against the spring force.

The spring force is determined by the spring constant \( k \), given as 120 N/m in this exercise. When the block is at its point of maximum compression, its kinetic energy \( KE \) has been converted into potential energy stored in the spring and work done against friction.

• The total work done against friction during compression is given by \( F_f \cdot x \), where \( x \) is the compression distance.
• The potential energy stored in the spring is expressed as \( \frac{1}{2}kx^2 \).

In this problem, the energy conservation equation setup would be \( KE = \frac{1}{2}kx^2 + F_fx \). By using this equation, you can solve for the compression distance \( x \) as the block comes to a stop, maximizing the compression of the spring.

Energy Conservation

Energy conservation is a fundamental principle that states energy cannot be created or destroyed, only transformed from one form to another. In this exercise, the block's initial kinetic energy is transformed into potential energy when compressing the spring.

Initially, the block has a kinetic energy \( KE \) equal to \( \frac{1}{2} m v^2 \), where \( m = 2.0 \) kg and \( v = 1.3 \) m/s. As the block compresses the spring, this energy is used for two things:

• Compressing the spring, storing energy as potential energy \( \frac{1}{2}kx^2 \).
• Doing work against the frictional force \( F_f \).

The energy conservation equation balances the block's initial kinetic energy with these two energy expenditures. If the static friction coefficient \( \mu_s \) is too low, the block will fail to remain stationary when the spring releases, leading to all the spring's potential energy reverting back to kinetic energy, causing the block to move once more. This interaction highlights the delicate balance of forces and energy transformations at play in dynamic systems.