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.

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

- \( m_1 = 100 \, \text{kg} \) is the mass of the block on the right side.
- \( h = 1 \, \text{m} \) is the distance the block travels.
- \( t = 6.3 \, \text{seconds} \) is the time it takes for the block to reach the ground.
- \( m_2 \) is the unknown mass of the block on the left side.
- \( g = 9.81 \, \text{m/s}^2 \) 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 \( m_1 \)

Since block \( m_1 \) starts from rest and takes \( t = 6.3 \) seconds to fall a height of \( h = 1 \) meter, we can use the kinematic equation:

\[
h = \frac{1}{2} a t^2
\]

Solving for the acceleration \( a \):

\[
1 = \frac{1}{2} a (6.3)^2
\]

\[
a = \frac{2 \times 1}{6.3^2} = \frac{2}{39.69} \approx 0.0504 \, \text{m/s}^2
\]

### Step 2: Newton's Second Law for Both Blocks

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

For \( m_1 \) (the block falling down):
\[
m_1 g - T = m_1 a
\]
where \( T \) is the tension in the string.

For \( m_2 \) (the block moving upwards):
\[
T - m_2 g = m_2 a
\]

### Step 3: Solving for \( m_2 \)

Now we have two equations with two unknowns (the tension \( T \) and the mass \( m_2 \)):

1. \( T = m_1 g - m_1 a \)
2. \( T = m_2 g + m_2 a \)

Equating these expressions for \( T \):

\[
m_1 g - m_1 a = m_2 g + m_2 a
\]

Substitute known values:

\[
100 \times 9.81 - 100 \times 0.0504 = m_2 \times 9.81 + m_2 \times 0.0504
\]

Simplify the equation:

\[
981 - 5.04 = 9.81 m_2 + 0.0504 m_2
\]

\[
975.96 = m_2 (9.81 + 0.0504)
\]

\[
975.96 = m_2 \times 9.8604
\]

Finally, solve for \( m_2 \):

\[
m_2 = \frac{975.96}{9.8604} \approx 99 \, \text{kg}
\]

### Final Answer:
The mass of the block on the left side is approximately \( \boxed{99 \, \text{kg}} \).

---

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!

Discover how to solve a pulley system problem where a 100kg block takes 6.3 seconds to fall 1 meter. Learn step-by-step calculations using Newton's Second Law and kinematic equations. Suitable for high school physics students.

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