Math Problem Statement

determine the acceleration of the system and the tension of each rope with sig figs

Determine the acceleration of the system and the tension of each rope with sig figs.

Solution

The image depicts a physics problem involving two masses connected by a pulley system. Here's what the setup involves:

  1. A mass 3m3m is placed on a horizontal surface.
  2. A second mass mm is hanging vertically.
  3. There is a coefficient of kinetic friction μ=0.25\mu = 0.25 between the 3m3m mass and the surface.
  4. We are tasked with calculating:
    • The acceleration of the system.
    • The tension in the rope.

Let’s analyze the forces and solve:


1. Define forces on each mass:

For the 3m3m block (horizontal):

  • Tension TT pulls the block to the right.
  • Frictional force fk=μFN=μ3mgf_k = \mu \cdot F_N = \mu \cdot 3m g, where FN=3mgF_N = 3m g is the normal force.
  • Net force: Fnet,3m=Tfk=T0.253mgF_{\text{net},3m} = T - f_k = T - 0.25 \cdot 3m g.

Newton's second law gives: T0.253mg=3maT - 0.25 \cdot 3m g = 3m a

For the mm block (vertical):

  • Gravitational force Fg=mgF_g = m g pulls it downward.
  • Tension TT opposes the downward pull.
  • Net force: Fnet,m=mgTF_{\text{net},m} = m g - T.

Newton's second law gives: mgT=mam g - T = m a


2. Set up equations and solve:

We now have two equations:

  1. T0.253mg=3maT - 0.25 \cdot 3m g = 3m a
  2. mgT=mam g - T = m a

Add these equations to eliminate TT: (T0.253mg)+(mgT)=3ma+ma(T - 0.25 \cdot 3m g) + (m g - T) = 3m a + m a

Simplify: mg0.253mg=4mam g - 0.25 \cdot 3m g = 4m a

mg(10.75)=4mam g (1 - 0.75) = 4m a

0.25mg=4ma0.25 m g = 4m a

a=0.25g4=g16a = \frac{0.25 g}{4} = \frac{g}{16}

Using g=9.8m/s2g = 9.8 \, \text{m/s}^2: a=9.8160.6125m/s2a = \frac{9.8}{16} \approx 0.6125 \, \text{m/s}^2


3. Find the tension TT:

Substitute a=g16a = \frac{g}{16} into the second equation mgT=mam g - T = m a: mgT=mg16m g - T = m \cdot \frac{g}{16}

T=mgmg16T = m g - m \cdot \frac{g}{16}

T=mg(1116)T = m g \left(1 - \frac{1}{16}\right)

T=mg1516T = m g \cdot \frac{15}{16}

Using g=9.8m/s2g = 9.8 \, \text{m/s}^2: T=m9.815169.19mNT = m \cdot 9.8 \cdot \frac{15}{16} \approx 9.19 m \, \text{N}


Final Results (with significant figures):

  1. Acceleration of the system: a0.613m/s2a \approx 0.613 \, \text{m/s}^2 (3 significant figures).
  2. Tension in the rope: T9.19mNT \approx 9.19 m \, \text{N} (3 significant figures).

Would you like a step-by-step diagram explanation or clarification?


Related Questions:

  1. What happens to the acceleration if the coefficient of friction increases?
  2. How does the tension vary if mm and 3m3m are interchanged?
  3. How would the system behave if the pulley had mass and friction?
  4. How does the angle of the surface affect the calculations if it's inclined?
  5. What is the work done by tension over a displacement of 5 meters?

Tip: Always double-check units and ensure the number of significant figures matches the input data!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Friction
Kinematics
Tension in ropes

Formulas

F = ma
Frictional force: f_k = μ * F_N
Net force equations for the blocks: T - μ(3mg) = 3ma and mg - T = ma

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 11-12