Question

A 1.00 -kg block is resting against a light, compressed spring at the bottom of a rough plane inclined at an angle of \(30.0^{\circ}\); the coefficient of kinetic friction between block and plane is \(\mu_{\mathrm{k}}=0.100 .\) Suppose the spring is compressed \(10.0 \mathrm{~cm}\) from its equilibrium length. The spring is then released, and the block separates from the spring and slides up the incline a distance of only \(2.00 \mathrm{~cm}\) beyond the spring's normal length before it stops. Determine a) the change in total mechanical energy of the system and b) the spring constant \(k\).

Short answer

Based on the given problem, the change in total mechanical energy when the block slides up the incline is 0.1018 J, and the spring constant is approximately 97.24 N/m.

Step-by-step solution

Gravitational potential energy

Initially, the block is at the bottom of the incline, so its gravitational potential energy is 0 J.

Kinetic energy

Initially, the block is at rest, so its kinetic energy is also 0 J.

Elastic potential energy

Since the spring is compressed 10 cm (\(0.1\,\text{m}\)), its elastic potential energy is given by: $$E_s = \frac{1}{2}kx^2$$ The initial energy state of the system is: $$E_i = 0\,\text{J} + 0\,\text{J} + \frac{1}{2}k(0.1^2)$$ #Step 2: Calculate the final energy state of the system# When the block stops, it doesn't have any kinetic energy. We need to find the gravitational potential energy and the elastic potential energy of the spring.

Gravitational potential energy

The vertical height (h) at the stopping point is given by: $$h = (0.1 + 0.02) \sin{(30^\circ)} = 0.06\,\text{m}$$ So, the gravitational potential energy is: $$PE = mgh = (1.00\,\text{kg})(9.8\,\text{m/s}^2)(0.06\,\text{m}) = 0.588\,\text{J}$$

Elastic potential energy

At the stopping point, the spring is at its equilibrium length, so there is no elastic potential energy. The final energy state of the system is: $$E_f = 0.588\,\text{J} + 0\,\text{J}$$ #Step 3: Calculate the change in total mechanical energy# Using the work-energy theorem, we can write: $$ΔME = W_{f} = E_f - E_i = 0.588\,\text{J} - \frac{1}{2}k(0.1^2)$$ Now we need to find the work done against friction.

Find the work done against friction

The normal force (N) can be found using: $$N = mg\cos{30^\circ}$$ Now we can find the friction force (f) using the coefficient of kinetic friction: $$f = \mu_{k}N = (0.100)(1.00\,\text{kg})(9.8\,\text{m/s}^2)\cos{30^\circ} = 0.848\,\text{N}$$ The work done against friction is the product of the friction force and the distance the block slides: $$W_{f} = fd = 0.848\,\text{N} \cdot 0.12\,\text{m} = 0.1018\,\text{J}$$ We can now solve for ΔME: $$ΔME = 0.1018\,\text{J}$$ #Step 4: Calculate the spring constant# We can substitute the expression for ΔME into the work-energy theorem equation: $$0.1018\,\text{J} = 0.588\,\text{J} - \frac{1}{2}k(0.1^2)$$ Now we solve for k: $$k = \frac{0.588\,\text{J} - 0.1018\,\text{J}}{0.005\,\text{m}^2} = 97.24\,\text{N/m}$$ So, the spring constant is approximately 97.24 N/m, and the change in total mechanical energy of the system is 0.1018 J.

Key concepts

Kinetic Energy

Kinetic energy is the energy of motion. In this exercise, the block initially rests against the compressed spring. This means that initially, its kinetic energy is zero because it is not moving. Kinetic energy is given by the formula:

• \( KE = \frac{1}{2}mv^2 \)

where \( m \) is mass and \( v \) is velocity.
The movement of the block starts once the spring releases the stored energy, converting some of its potential energy to kinetic energy, giving the block motion as it slides up the incline.
However, when the block comes to a stop 2.00 cm beyond the spring’s normal length, its kinetic energy is again 0 J, as its velocity is zero.

Gravitational Potential Energy

Gravitational potential energy depends on an object's height above a reference point. It is expressed as:

• \( PE = mgh \)

where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above the reference point.
In this exercise, when the block is at the bottom of the incline, its gravitational potential energy is initially 0 J, as it measures height from this point.
As the block climbs the slope, it gains gravitational potential energy proportional to its height on the inclined plane. At its highest point, the block's gravitational potential energy is 0.588 J, calculated using the inclination angle and the vertical height.
This rise in potential energy demonstrates how energy is redistributed within the system.

Elastic Potential Energy

Elastic potential energy is stored in objects when they are compressed or stretched. In this context with the spring, it’s defined by the equation:

• \( E_s = \frac{1}{2}kx^2 \)

where \( k \) is the spring constant and \( x \) is the amount of compression or extension from the spring's equilibrium position.
Initially, the spring is compressed by 10 cm, leading to elastic potential energy being stored in it. When released, this stored energy is converted into kinetic energy, moving the block.
When the block is moving back to its natural position, the elastic potential energy reduces to zero, as the spring returns to its original length at the highest point reached by the block.
Understanding elastic potential energy helps explain the motion initiated and stopped by the spring's force.

Work-Energy Theorem

The work-energy theorem provides a relationship between the work done on an object and its change in energy states. This theorem states:

• \( ΔKE = W \)

where \( ΔKE \) is the change in kinetic energy and \( W \) is the work done on the object.
In the exercise, the total mechanical energy of the block-spring system changes due to work done by various forces like gravity and friction.
The work done against friction accounts for the energy loss in the system when the block slides along the incline. This energy is not destroyed but converted mostly to heat, showing a decrease in the block's mechanical energy.
By equating changes in energy states and the work done, one can solve for unknowns such as the spring constant, calculating the precise energy conversions within the system.