91Ó°ÊÓ

A box of bananas weighing 40.0 \(\mathrm{N}\) rests on a horizontal surface. The coefficient of static friction between the box and the sur- face is \(0.40,\) and the coefficient of kinetic friction is 0.20 . (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 \(\mathrm{N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of \(18.0 \mathrm{N},\) what is the magnitude of the friction force and what is the box's acceleration?

Step by step solution

01

Determine Static Friction (No Applied Force)

When no horizontal force is applied, the box remains at rest due to static friction. The static friction force equals the gravitational force on the box, which is 40.0 N, because it perfectly balances any potential movement. Thus, the friction force is also 0 N since no pushing force exists.

02

Calculate Static Friction Against Applied 6 N Force

Static friction will adjust up to its maximum limit to prevent motion. Apply a horizontal force of 6.0 N, less than the maximum static friction force, meaning static friction opposes this force fully, resulting in a friction force of 6.0 N.

03

Determine Minimum Force to Initiate Movement

The force required to overcome maximum static friction is obtained by multiplying the coefficient of static friction (0.40) by the normal force (40.0 N). Therefore, the force is: \[ f_s = 0.40 \times 40.0 = 16.0 \text{ N} \] Hence, a minimum of 16.0 N is needed to start moving the box.

04

Calculate Minimum Force for Constant Velocity

Once in motion, kinetic friction opposes motion. Calculate this force by multiplying the coefficient of kinetic friction (0.20) by the normal force (40.0 N). So, the force is: \[ f_k = 0.20 \times 40.0 = 8.0 \text{ N} \] This force of 8.0 N is necessary to maintain constant velocity.

05

Assess Friction Force and Acceleration with 18 N Applied Force

Kinetic friction (8.0 N) opposes the 18.0 N force. Calculate acceleration using Newton's second law. The net force is \[ F = 18.0 \text{ N} - 8.0 \text{ N} = 10.0 \text{ N} \]Using mass from weight \[ m = \frac{40.0}{9.8} \approx 4.08 \text{ kg} \]The acceleration is \[ a = \frac{10.0}{4.08} \approx 2.45 \text{ m/s}^2 \] Friction force is 8.0 N while the box accelerates at 2.45 m/s².

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Friction

Static friction acts on objects that aren't in motion relative to a surface. It is what keeps our box of bananas securely stationary until a force strong enough is applied to overcome it.

Static friction will adjust itself to oppose any external force trying to initiate movement, up to its maximum limit. In our case, the maximum static friction force before the bananas start to slide is:

• Calculated as the product of the coefficient of static friction (0.40) and the normal force (equal to the weight of the box, 40.0 N)

Thus, the maximum static friction is 16.0 N. Only a force greater than this will set the bananas in motion.

If the external force is less than 16.0 N, static friction will balance it out completely, making the net force zero, and hence, no movement occurs. For instance, a 6.0 N push won't move the box; the friction force matches this push at 6.0 N.

Kinetic Friction

When the bananas start to slide, static friction gives way to kinetic friction. This type of friction acts between surfaces in relative motion.

Kinetic friction usually has a smaller coefficient than static friction, indicating it's easier to keep an object moving once it has started. For our box, the kinetic friction coefficient is 0.20. The kinetic friction force is:

• Found by multiplying the coefficient of kinetic friction (0.20) by the normal force (40.0 N)

Resulting in a kinetic friction force of 8.0 N.

To maintain a constant velocity, any applied force must exactly balance this kinetic friction force. Thus, a continuous 8.0 N of horizontal force will keep the box sliding at a constant speed.

Newton's Second Law of Motion

Newton's Second Law relates force, mass, and acceleration in the equation: \[F = ma\]Here, force is the net external force applied, mass is the object's mass, and acceleration is how much speed changes over time.

In the final case of the monkey applying an 18.0 N force, we use the law to calculate acceleration. First, kinetic friction opposes the movement with an 8.0 N force, leading to a net force:

• Net force: 18.0 N - 8.0 N = 10.0 N

Next, calculate the box's mass from its weight using gravity: \( \ m = \frac{W}{g} = \frac{40.0}{9.8} \approx 4.08 \text{ kg} \\)

Finally, use Newton's Second Law to find acceleration: \( \ a = \frac{F}{m} = \frac{10.0}{4.08} = 2.45 \text{ m/s}^2 \) This means the excess force not used in overcoming friction causes the box to accelerate at 2.45 m/s².