Problem 13 - University Physics with Modern Physics

Question

A 10.0 -kg microwave oven is pushed 8.00 \(\mathrm{m}\) up the sloping surface of a loading ramp inclined at an angle of \(36.9^{\circ}\) above the horizontal, by a constant force \(\vec{F}\) with a magnitude 110 \(\mathrm{N}\) 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 \(\vec{F} ?\) (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 \(\Sigma \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\boldsymbol{a}}\) to calculate the acceleration of the oven. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 8.00 \(\mathrm{m}\) . From this, compute the increase in the oven's kinetic energy, and compare it to the answer you got in part (d).

Step-by-Step Solution

Answer

(a) 880 J, (b) -156.8 J, (c) 470.4 J, (d) 252.8 J, (e) Acceleration = 3.16 m/s², Speed = 7.11 m/s, Increase in KE = 252.8 J.

Step 1 - Work Done by Force F

The work done by a force is calculated by the formula \( W = F \cdot d \cdot \cos(\theta) \). Here, the force \( \vec{F} = 110 \, \text{N} \) acts parallel to the ramp, so \( \theta = 0^\circ \).\[ W = 110 \, \text{N} \times 8.00 \, \text{m} \times \cos(0^\circ) = 880 \, \text{J} \]Therefore, the work done by the force \( \vec{F} \) is 880 J.

Step 2 - Work Done by Friction

The frictional force is calculated using \( f_k = \mu_k \cdot N \), where \( \mu_k = 0.250 \) and \( N \) is the normal force.The normal force \( N \) is given by \( N = mg \cdot \cos(36.9^\circ) \), where \( m = 10.0 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).\[ N = 10.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(36.9^\circ) \approx 78.4 \, \text{N} \]Then:\[ f_k = 0.250 \times 78.4 \, \text{N} = 19.6 \, \text{N} \]The work done by friction is \( W_f = f_k \times d \times \cos(180^\circ) \) (friction acts opposite to displacement):\[ W_f = 19.6 \, \text{N} \times 8.00 \, \text{m} \times (-1) = -156.8 \, \text{J} \]Thus, the work done by the friction force is \(-156.8 \, \text{J}\).

Step 3 - Increase in Potential Energy

The increase in gravitational potential energy is given by \( \Delta U = mgh \), where \( h = d \sin(\theta) \).\[ h = 8.00 \, \text{m} \times \sin(36.9^\circ) \approx 4.80 \, \text{m} \]\[ \Delta U = 10.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 4.80 \, \text{m} = 470.4 \, \text{J} \]The increase in potential energy is 470.4 J.

Step 4 - Increase in Kinetic Energy (Part D)

According to the work-energy principle, the total work done is equal to the change in kinetic energy.Total work done \( = W_{\vec{F}} + W_f = 880 \, \text{J} + (-156.8 \, \text{J}) = 723.2 \, \text{J} \).Since \( \Delta KE = \text{Total Work} - \Delta U \), we compute:\[ \Delta KE = 723.2 \, \text{J} - 470.4 \, \text{J} = 252.8 \, \text{J} \]The increase in the oven's kinetic energy is 252.8 J.

Step 5 - Calculate Acceleration

Use Newton's second law: \( \Sigma F = ma \).Net force along the ramp is \( F_{\text{net}} = (F - f_k - mg \sin(36.9^\circ)) \):Calculate \( mg \sin(36.9^\circ) \):\[ mg \sin(36.9^\circ) = 10.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(36.9^\circ) \approx 58.8 \, \text{N} \]\[ F_{\text{net}} = 110 \, \text{N} - 19.6 \, \text{N} - 58.8 \, \text{N} = 31.6 \, \text{N} \]Now solve for acceleration \( a \):\[ a = \frac{F_{\text{net}}}{m} = \frac{31.6 \, \text{N}}{10.0 \, \text{kg}} = 3.16 \, \text{m/s}^2 \]Thus, the acceleration of the oven is \( 3.16 \, \text{m/s}^2 \).

Step 6 - Calculate Speed and Increase in Kinetic Energy (Part E)

Using \( v^2 = u^2 + 2as \) (where \( u = 0 \) because the oven is initially at rest), to find speed \( v \):\[ v^2 = 0 + 2(3.16 \, \text{m/s}^2)(8.00 \, \text{m}) \]\[ v = \sqrt{50.56} \approx 7.11 \, \text{m/s} \]Now compute \( \Delta KE = \frac{1}{2}mv^2 \):\[ \Delta KE = \frac{1}{2}(10.0 \, \text{kg})(7.11 \, \text{m/s})^2 \approx 252.8 \, \text{J} \]The increase in kinetic energy is 252.8 J, the same as part (d), confirming consistency.

Key Concepts

Kinetic Friction

Kinetic friction is a type of friction that acts between moving surfaces. It's important to understand when dealing with objects sliding down a ramp.

• The kinetic frictional force, denoted as \( f_k \), opposes the motion of the object.
• It is calculated using the formula \( f_k = \mu_k \cdot N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.

In this exercise, the coefficient of kinetic friction is given as 0.250. The normal force is calculated as \( N = mg \cdot \cos(36.9^\circ) \), where the weight of the microwave oven contributes to the normal force on the inclined surface.
After finding the normal force to be approximately 78.4 N, we calculate the kinetic frictional force as 19.6 N. This frictional force does negative work because it acts in the opposite direction of the motion.
The work done by friction is computed by multiplying the frictional force, the distance, and \( \cos(180^\circ) \) (since the direction is opposite), resulting in \(-156.8 \) J of work done by friction.

Potential Energy

Potential energy is the energy stored in an object due to its position relative to some zero position; in this case, the height on a ramp. Gravitational potential energy increases as you move an object against gravity.

• The gravitational potential energy \( U \) is given by \( \Delta U = mgh \), where \( h \) is the height gained.
• In inclined planes, the height \( h \) can be determined using trigonometry. It is \( h = d \sin(\theta) \).

For this exercise, the height gained by the microwave oven after moving up the ramp is calculated using \( \sin(36.9^\circ) \), resulting in approximately 4.80 meters of height.
Using the mass (10.0 kg) and gravitational acceleration (9.8 m/s²), the increase in potential energy is calculated as 470.4 J.
This increase in potential energy tells us how much energy was required to raise the microwave to the new position above ground level.

Newton's Second Law

Newton's Second Law establishes the relationship between force, mass, and acceleration in a simple mathematical way. It is expressed as \( \Sigma \vec{F} = m\vec{a} \).

• This law states that the acceleration \( a \) of an object is directly proportional to the net force \( F_{\text{net}} \) acting on the object and inversely proportional to the object's mass \( m \).
• The net force can be calculated as the sum of all the forces acting on the object along the direction of the motion.

In the context of the exercise, several forces act along the ramp:- The applied force \( F = 110 \) N- Friction force \( f_k = 19.6 \) N- The component of the gravitational force along the ramp \( mg \sin(36.9^\circ) \), calculated as approximately 58.8 N
The net force \( F_{\text{net}} = 110 \text{ N} - 19.6 \text{ N} - 58.8 \text{ N} = 31.6 \text{ N} \) leads to an acceleration calculated using \( a = \frac{F_{\text{net}}}{m} \), giving us \( a = 3.16 \ \text{m/s}^2 \).
This acceleration, along with the kinematics equations, helps in determining the speed of the microwave after moving 8.00 m up the ramp.