**Instructions** Click 'start' in the following Active Figure depicting static and kinetic frictional forces with a given applied force to a trash can to complete the exercise.**EXPLORE** A trash can with a mass of 11 kg has a coefficient of static friction of \( \mu_s = 0.88 \) and a coefficient of kinetic friction of \( \mu_k = 0.40 \). (The Active Figure applies to a similar situation, but the values of the mass, coefficient of static friction, and coefficient of kinetic friction may be different. So the values it gives may not be the same values you obtain from your calculations.)(A) What is the maximum horizontal force \( \vec{F} \) that can be exerted without moving the trash can?(B) Suppose this force is just barely exceeded. Determine the acceleration of the trash can as it moves.**CONCEPTUALIZE** The trash can is being pushed near its base, but the applied force is balanced by the force of static friction. The maximum force of static friction is determined by the coefficient of static friction and the normal force that the ground exerts on the trash can._Display in a New Window_ **CATEGORIZE** We seek the largest horizontal force \( \vec{F} \) that can be applied without moving the trash can. Examine when it is possible to balance the forces to have no acceleration; this is a problem of balanced forces and equilibrium. **ANALYZE****(A) What is the maximum horizontal force \(\vec{F}\) that can be exerted without moving the trash can?**The surfaces are extremely rough, and only a fraction of the surfaces is in close contact, but we make the assumption that the relation between the friction force and normal force applies.The maximum possible force of static friction is \[ F_{s, \text{max}} = \mu_s n \]where \( n \) is the magnitude of the normal force exerted by one surface on the other. For an applied force that is too small to move the trash can, the force of static friction balances the applied force and both have the same magnitude which, because \( n = mg \), is equal to\[ F = F_{s, \text{max}} = \mu_s mg \]where \( \mu_s \) is the coefficient of static friction. This gives\[ F = F_{s, \text{max}} = \underline{\hspace{3cm}} \, \text{N} \]**(B) Determine the acceleration of the trash can as it moves.**For a force \( F \) that is just slightly larger than \( F_{s, \text{max}} \), the trash can accelerates. The applied force can still be taken as the same, but the friction force is given by the coefficient of kinetic friction \( \mu_k \) times the normal force \( n \) instead of the coefficient of static friction \( \mu_s \) times the normal force \( n \):\[ F_k = \mu_k mg \]The net force that gives the mass times the acceleration is the difference of the applied and kinetic friction forces:\[ \sum \vec{F}_x = F_{s, \text{max}} - \mu_k mg = \mu_s mg - \mu_k mg = (\mu_s - \mu_k) mg = ma_x \]This can be solved to give the acceleration:\[ a_x = (\mu_s - \mu_k) g = \underline{\hspace{3cm}} \, \text{m/s}^2 \]**FINALIZE**If the applied force is suddenly removed from the moving trash can, the force of kinetic friction continues to act with no applied force as long as the trash can is moving. Consider how this affects the motion of

Name: **Instructions** Click 'start' in the following Active Figure depict
Uploaded: 2024-03-20T06:57:57.000Z
Channel: Bartleby
Description: **Instructions** Click 'start' in the following Active Figure depicting static and kinetic frictional forces with a given applied force to a trash can to complete the exercise.**EXPLORE** A trash can with a mass of 11 kg has a coefficient of stati