Problem 13 - University Physics with Modern Physics

Question

A 10.0-kg microwave oven is pushed 6.00 m up the sloping surface of a loading ramp inclined at an angle of 36.9\(^\circ\) above the horizontal, by a constant force \(\overrightarrow{F}\) with a magnitude 110 N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250. (a) What is the work done on the oven by the force \(\overrightarrow{F}\)? (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy. (e) Use \(\sum \overrightarrow{F} = m\overrightarrow{a}\) to calculate the oven's acceleration. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after the oven has traveled 6.00 m. From this, compute the increase in the oven's kinetic energy, and compare it to your answerfor part (d).

Step-by-Step Solution

Answer

(a) 660 J; (b) -117.6 J; (c) 352.8 J; (d) 189.6 J; (e) 6.16 m/s velocity.

Step 1: Calculate Work Done by Force F

The work done by the force \( \overrightarrow{F} \) is given by the formula:\[ W = F \cdot d \cdot \cos(\theta) \]Where \( F = 110 \, \text{N} \), the distance \( d = 6.00 \, \text{m} \), and \( \theta = 0^\circ \) because the force is parallel to the ramp. Thus, the work done is:\[ W = 110 \, \text{N} \times 6.00 \, \text{m} \times \cos(0^\circ) = 660 \, \text{J} \]

Step 2: Calculate Work Done by Friction

The friction force can be computed using the formula:\[ f_k = \mu_k \cdot N \]Where \( \mu_k = 0.250 \) and \( N = mg \cos(\alpha) \), with \( \alpha = 36.9^\circ \). First, calculate the normal force:\[ N = 10.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(36.9^\circ) = 78.4 \, \text{N} \]Now compute the friction force:\[ f_k = 0.250 \times 78.4 \, \text{N} = 19.6 \, \text{N} \]Then, the work done by friction is:\[ W_f = f_k \cdot d \cdot \cos(180^\circ) = 19.6 \, \text{N} \times 6.00 \, \text{m} \times (-1) = -117.6 \, \text{J} \]

Step 3: Calculate Increase in Potential Energy

The increase in potential energy \( \Delta U \) is given by:\[ \Delta U = mgh \]Where \( h = d \sin(\alpha) \). Thus,\[ h = 6.00 \, \text{m} \times \sin(36.9^\circ) = 3.60 \, \text{m} \]Therefore, the increase in potential energy is:\[ \Delta U = 10.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3.60 \, \text{m} = 352.8 \, \text{J} \]

Step 4: Calculate Increase in Kinetic Energy

The increase in kinetic energy \( \Delta KE \) is the net work done on the oven:\[ \Delta KE = W - |W_f| - \Delta U \]Substituting the values:\[ \Delta KE = 660 \, \text{J} - 117.6 \, \text{J} - 352.8 \, \text{J} = 189.6 \, \text{J} \]

Step 5: Find Acceleration Using \( \sum \overrightarrow{F} = m\overrightarrow{a} \)

Apply Newton's second law along the ramp:\[ F - f_k - mg \sin(\alpha) = ma \]Substitute the values:\[ 110 \, \text{N} - 19.6 \, \text{N} - 10.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(36.9^\circ) = 10.0 \, \text{kg} \times a \]Simplifying:\[ 110 \, \text{N} - 19.6 \, \text{N} - 58.8 \, \text{N} = 10.0 \, \text{kg} \times a \]\[ 31.6 \, \text{N} = 10.0 \, \text{kg} \times a \]\[ a = 3.16 \, \text{m/s}^2 \]

Step 6: Calculate Final Speed and Verify Kinetic Energy

Use the kinematic equation:\[ v^2 = u^2 + 2ad \]Assume initial velocity \( u = 0 \):\[ v^2 = 2 \times 3.16 \, \text{m/s}^2 \times 6.00 \, \text{m} \]\[ v^2 = 37.92 \]\[ v \approx 6.16 \, \text{m/s} \]The increase in kinetic energy is then:\[ \Delta KE = \frac{1}{2} mv^2 = \frac{1}{2} \times 10.0 \, \text{kg} \times (6.16 \, \text{m/s})^2 = 189.8 \, \text{J} \]This value matches the kinetic energy calculated in Step 4.

Key Concepts

Kinetic Friction

When we talk about kinetic friction, we are referring to the force that opposes the motion of two surfaces sliding past each other. In this exercise, the microwave oven is pushed up a ramp, causing kinetic friction to act between the oven and the ramp.
This frictional force can be calculated by multiplying the coefficient of kinetic friction (\( \mu_k = 0.250 \)) by the normal force exerted on the oven. The normal force is perpendicular to the ramp's surface and can be determined using the formula:

• \( N = mg \cos(\alpha) \)

where \( \alpha \) is the angle of the ramp (36.9° in this case).
After calculating the normal force, we find:

• \( f_k = \mu_k \times N \)
• \( f_k = 19.6 \text{ N} \)

This frictional force is crucial because it does negative work on the oven, meaning it takes away energy as the oven is pushed up the ramp, calculated as \( -117.6 \text{ J} \). It's essential to consider this when determining the oven's total energy changes.

Potential Energy

Potential energy refers to the stored energy in an object due to its position relative to some zero position. In this situation, as the microwave oven moves up the ramp, its potential energy increases because it's gaining height. This increase can be calculated by the formula:

• \( \Delta U = mgh \)

Here, \( m \) is the mass of the oven (10 kg), \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the vertical height gained, which can be calculated using the sine of the ramp angle:

• \( h = d \sin(\alpha) \)
• \( h = 3.60 \text{ m} \)

Thus, the increase in potential energy is \( 352.8 \text{ J} \).
This potential energy gain is important, as it represents the energy required to lift the oven up to its new height, against the pull of gravity.

Newton's Second Law

Newton's Second Law of Motion is a fundamental principle that describes how the motion of an object changes when it is subject to forces. In simpler terms, it tells us that an object will accelerate if a net force is applied to it, and this acceleration is directly proportional to the force and inversely proportional to the mass:\[ \sum \overrightarrow{F} = m \overrightarrow{a} \]To calculate the acceleration of the oven, we must consider all forces acting along the ramp: the pushing force (110 N), the opposing frictional force (19.6 N), and the component of gravitational force pulling the oven down the ramp (\( mg \sin(\alpha) \)). Substituting these values, we reach:

• \( a = 3.16 \text{ m/s}^2 \)

This acceleration tells us how fast the oven's velocity changes as it moves up the ramp.
Finally, with the acceleration determined, the velocity and the increase in kinetic energy after traveling 6 meters can be validated, providing key insights into how energy principles and motion laws operate together in this context.