Math Problem Statement

There are two blocks in a pulley. One of them is a 100kg block that is 1 meter above the ground. This block takes 6.3 seconds to reach the ground after being released from rest. What is the mass of the block on the left? The pulley is massless and frictionless.

Solution

We can solve this problem using Newton's Second Law and the principles of mechanics for pulley systems. Let's denote the following:

  • m1=100kgm_1 = 100 \, \text{kg} is the mass of the block on the right side.
  • h=1mh = 1 \, \text{m} is the distance the block travels.
  • t=6.3secondst = 6.3 \, \text{seconds} is the time it takes for the block to reach the ground.
  • m2m_2 is the unknown mass of the block on the left side.
  • g=9.81m/s2g = 9.81 \, \text{m/s}^2 is the acceleration due to gravity.
  • Assume the pulley is massless and frictionless, so the tension is the same on both sides.

Step 1: Equation of Motion for m1m_1

Since block m1m_1 starts from rest and takes t=6.3t = 6.3 seconds to fall a height of h=1h = 1 meter, we can use the kinematic equation:

h=12at2h = \frac{1}{2} a t^2

Solving for the acceleration aa:

1=12a(6.3)21 = \frac{1}{2} a (6.3)^2

a=2×16.32=239.690.0504m/s2a = \frac{2 \times 1}{6.3^2} = \frac{2}{39.69} \approx 0.0504 \, \text{m/s}^2

Step 2: Newton's Second Law for Both Blocks

Now, apply Newton's Second Law to each block:

For m1m_1 (the block falling down): m1gT=m1am_1 g - T = m_1 a where TT is the tension in the string.

For m2m_2 (the block moving upwards): Tm2g=m2aT - m_2 g = m_2 a

Step 3: Solving for m2m_2

Now we have two equations with two unknowns (the tension TT and the mass m2m_2):

  1. T=m1gm1aT = m_1 g - m_1 a
  2. T=m2g+m2aT = m_2 g + m_2 a

Equating these expressions for TT:

m1gm1a=m2g+m2am_1 g - m_1 a = m_2 g + m_2 a

Substitute known values:

100×9.81100×0.0504=m2×9.81+m2×0.0504100 \times 9.81 - 100 \times 0.0504 = m_2 \times 9.81 + m_2 \times 0.0504

Simplify the equation:

9815.04=9.81m2+0.0504m2981 - 5.04 = 9.81 m_2 + 0.0504 m_2

975.96=m2(9.81+0.0504)975.96 = m_2 (9.81 + 0.0504)

975.96=m2×9.8604975.96 = m_2 \times 9.8604

Finally, solve for m2m_2:

m2=975.969.860499kgm_2 = \frac{975.96}{9.8604} \approx 99 \, \text{kg}

Final Answer:

The mass of the block on the left side is approximately 99kg\boxed{99 \, \text{kg}}.


Would you like more details on any of these steps or another explanation?

Here are 5 related questions to deepen your understanding:

  1. How would the solution change if the pulley had mass?
  2. How is the tension in the string affected if there is friction in the pulley?
  3. Can you solve for the acceleration using energy principles instead of Newton's laws?
  4. What happens to the system if the mass of the left block is greater than the right block?
  5. How would the time to reach the ground change if the height were doubled?

Tip: Always check if the acceleration makes sense for pulley systems—if it’s too large or too small, double-check your setup!

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Math Problem Analysis

Mathematical Concepts

Kinematics
Newton's Second Law
Pulley System Dynamics

Formulas

h = (1/2) a t^2
Newton's Second Law: F = ma
Tension equations: T = m1g - m1a and T = m2g + m2a

Theorems

Newton's Second Law
Kinematic equations for uniformly accelerated motion

Suitable Grade Level

Grades 11-12 (High School Physics)