Math Problem Statement

The coefficient of static friction is 0.60 between the two blocks in figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F⃗ causes both blocks to cross a distance of 5.0 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

Solution

To solve this problem, we need to analyze the forces acting on the system and determine the minimum time required for the motion, given that the top block must not slide on the lower block. Here's a step-by-step breakdown:

Step 1: Understanding the forces involved

  • Static friction between the two blocks (top and lower block) resists the relative motion between them. The maximum static friction force, fsf_s, is given by: fs=μsm1gf_s = \mu_s m_1 g where μs=0.60\mu_s = 0.60 is the coefficient of static friction, m1m_1 is the mass of the top block, and g=9.8m/s2g = 9.8 \, \text{m/s}^2 is the acceleration due to gravity.

  • Kinetic friction between the lower block and the floor resists the motion of the system. The kinetic friction force, fkf_k, is given by: fk=μk(m1+m2)gf_k = \mu_k (m_1 + m_2) g where μk=0.20\mu_k = 0.20 is the coefficient of kinetic friction, and m2m_2 is the mass of the lower block.

Step 2: Forces acting on the system

The force FF must accelerate both blocks together across a distance of 5.0 m without causing the top block to slide on the lower block. This means the acceleration of the system, aa, must be such that the static friction force is not exceeded.

  1. The total mass of the system is m1+m2m_1 + m_2.

  2. The net force required to accelerate the system is: Ffk=(m1+m2)aF - f_k = (m_1 + m_2) a where FF is the applied force and fkf_k is the kinetic friction.

  3. For the top block not to slide, the static friction must provide enough force to cause the same acceleration aa: fsm1af_s \geq m_1 a Substituting fs=μsm1gf_s = \mu_s m_1 g, we get: μsm1gm1a\mu_s m_1 g \geq m_1 a This gives the maximum acceleration aa: aμsga \leq \mu_s g Substituting μs=0.60\mu_s = 0.60 and g=9.8m/s2g = 9.8 \, \text{m/s}^2, we find: a0.60×9.8=5.88m/s2a \leq 0.60 \times 9.8 = 5.88 \, \text{m/s}^2

Step 3: Kinematic equation to find minimum time

The blocks start from rest and move a distance of 5.0 m. The kinematic equation that relates distance dd, acceleration aa, and time tt is: d=12at2d = \frac{1}{2} a t^2 Solving for time tt: t=2dat = \sqrt{\frac{2d}{a}} Substitute d=5.0md = 5.0 \, \text{m} and a=5.88m/s2a = 5.88 \, \text{m/s}^2: t=2×5.05.88=1.70071.30secondst = \sqrt{\frac{2 \times 5.0}{5.88}} = \sqrt{1.7007} \approx 1.30 \, \text{seconds}

Conclusion:

The least amount of time in which the motion can be completed without the top block sliding on the lower block is approximately 1.30 seconds.


Would you like more details on any part of the solution? Here are some follow-up questions to expand your understanding:

  1. What happens if the coefficient of static friction is lower than 0.60?
  2. How would the answer change if the blocks start with an initial velocity instead of from rest?
  3. What effect would a higher coefficient of kinetic friction have on the minimum time?
  4. How can we calculate the force FF needed to achieve this motion?
  5. What if the distance traveled were different, say 10 meters instead of 5 meters?

Tip: Always ensure the static friction is enough to prevent slipping, especially when dealing with multiple bodies in contact during motion.

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Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Friction
Kinematics

Formulas

f_s = μ_s * m_1 * g
f_k = μ_k * (m_1 + m_2) * g
F - f_k = (m_1 + m_2) * a
d = (1/2) * a * t^2
t = sqrt((2 * d) / a)

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 11-12