Questions: 4. Telesi z masama 1.5 kg in 2 kg sta povezani z lahko vrvjo in ležita na vodoravni gladki podlagi. S kolikšno največjo silo smemo še vleči drugo telo proč od prvega, da se vrv ne pretrga? Vrv še prenese silo 60 N. Kako pa je, če je masa vrvi 0.1 kg ?

Solution Steps

We have two masses, \( m_1 = 1.5 \, \text{kg} \) and \( m_2 = 2 \, \text{kg} \), connected by a light rope on a frictionless horizontal surface. We need to find the maximum force that can be applied to the second mass without breaking the rope, which can withstand a maximum tension of 60 N. We will also consider the case where the rope has a mass of 0.1 kg.

First, consider the system where the rope is massless. The maximum tension the rope can handle is 60 N. When a force \( F \) is applied to the second mass, the tension in the rope will be equal to the force required to accelerate the first mass \( m_1 \).

Using Newton's second law for \( m_1 \): \[ T = m_1 \cdot a \] where \( T = 60 \, \text{N} \) is the maximum tension and \( a \) is the acceleration of the system.

The total mass of the system is \( m_1 + m_2 = 1.5 \, \text{kg} + 2 \, \text{kg} = 3.5 \, \text{kg} \).

Using Newton's second law for the entire system: \[ F = (m_1 + m_2) \cdot a \]

Substituting \( a = \frac{T}{m_1} \) into the equation for \( F \): \[ F = (m_1 + m_2) \cdot \frac{T}{m_1} \]

Substitute the known values: \[ F = (3.5) \cdot \frac{60}{1.5} = 140 \, \text{N} \]

Now, consider the rope has a mass of 0.1 kg. The total mass of the system becomes \( m_1 + m_2 + m_{\text{rope}} = 3.6 \, \text{kg} \).

The tension in the rope will still be 60 N, but now the acceleration \( a \) is given by: \[ a = \frac{T}{m_1 + \frac{m_{\text{rope}}}{2}} \]

The force \( F \) applied to the system is: \[ F = (m_1 + m_2 + m_{\text{rope}}) \cdot a \]

Substitute the known values: \[ a = \frac{60}{1.5 + \frac{0.1}{2}} = \frac{60}{1.55} \approx 38.7097 \, \text{m/s}^2 \]

\[ F = 3.6 \cdot 38.7097 \approx 139.355 \, \text{N} \]

Final Answer