A ketchup bottle of mass 0.55 kg, with an initial velocity of 2.8 m/s, slides on a counter and slows due to friction, stopping after 0.10 m. Rigorously determine the magnitude and... A ketchup bottle of mass 0.55 kg, with an initial velocity of 2.8 m/s, slides on a counter and slows due to friction, stopping after 0.10 m. Rigorously determine the magnitude and direction of the average frictional force, applying the work-energy theorem.
Understand the Problem
The question asks to use the work-energy theorem to find the magnitude and direction of the average frictional force acting on a ketchup bottle as it slides and comes to a stop. The work-energy theorem directly relates the work done by all forces acting on an object to the change in the object's kinetic energy. In this case, the work is done by friction, and the change in kinetic energy is from the initial velocity to zero velocity.
Answer
The magnitude of the average frictional force is approximately $2.61 \text{ N}$, and the direction is opposite to the direction of motion.
Answer for screen readers
The magnitude of the average frictional force is approximately $2.61 \text{ N}$, and the direction is opposite to the direction of motion.
Steps to Solve
- Define the Work-Energy Theorem
The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. Mathematically, this is expressed as:
$$W_{net} = \Delta KE = KE_f - KE_i$$
Where $W_{net}$ is the net work done, $KE_f$ is the final kinetic energy, and $KE_i$ is the initial kinetic energy.
- Express Work Done by Friction
The work done by friction is given by:
$W_{friction} = -f_d \cdot d$
where $f_d$ is the magnitude of the average frictional force, and $d$ is the distance over which the force acts. The negative sign indicates that the frictional force opposes the motion.
- Express Initial and Final Kinetic Energies
The kinetic energy is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity.
The initial kinetic energy $KE_i$ is:
$KE_i = \frac{1}{2}mv_i^2$
where $v_i$ is the initial velocity. Since the bottle comes to rest, the final kinetic energy $KE_f$ is zero.
$KE_f = 0$
- Apply the Work-Energy Theorem
Substitute the expressions for work and kinetic energies into the work-energy theorem:
$-f_d \cdot d = 0 - \frac{1}{2}mv_i^2$
- Solve for the Frictional Force $f_d$
Rearrange the equation to solve for $f_d$:
$f_d = \frac{mv_i^2}{2d}$
- Plug in the Given Values
Given $m = 0.42 \text{ kg}$, $v_i = 3.7 \text{ m/s}$, and $d = 1.1 \text{ m}$, substitute these values into the equation:
$f_d = \frac{0.42 \text{ kg} \cdot (3.7 \text{ m/s})^2}{2 \cdot 1.1 \text{ m}}$
$f_d = \frac{0.42 \cdot 13.69}{2.2} = \frac{5.7498}{2.2} \approx 2.61 \text{ N}$
- Determine the Direction of the Frictional Force
The frictional force opposes the motion. Since the bottle is sliding to the right (assuming positive direction), the frictional force acts to the left (negative direction).
The magnitude of the average frictional force is approximately $2.61 \text{ N}$, and the direction is opposite to the direction of motion.
More Information
The work-energy theorem provides a simple way to relate work and kinetic energy, avoiding the need to directly calculate acceleration and forces using Newton's laws. This approach is particularly useful when the forces are not constant or the motion is complex.
Tips
A common mistake is forgetting the negative sign in the work done by friction. Friction always opposes the motion, so the work done by friction is negative, reducing the kinetic energy. Another error is using the incorrect units, ensure all units are in the SI system (kg, m, s). A final mistake would be in misinterpreting the direction of the frictional force.
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