In figure shown on the right, the mass of the trolley is 100 kg, and it can move without friction on the horizontal floor. Its length is 12m. The mass of the girl is 50 kg. Friction exists between the shoes of the girl and the trolley''s upper surface, with µ = 1/3. The girl can run with a maximum speed = 9 m/s on the surface of the trolley, with respect to the surface. At t = 0 the girl starts running from rest to the right. The trolley was initially stationary. (g = 10 m/s2) The minimum time in which the girl can acquire her maximum speed, for no slipping, is____

To solve this problem, we need to analyze the motion of both the girl and the trolley. The girl is running on the trolley, and we need to ensure that she does not slip. This means we must consider the forces acting on her and the resulting motion of the trolley. Let's break it down step by step.

Understanding the Forces at Play

The girl has a mass of 50 kg and can run at a maximum speed of 9 m/s relative to the trolley. The trolley has a mass of 100 kg and can move without friction. The coefficient of friction between the girl’s shoes and the trolley is given as µ = 1/3.

Calculating the Maximum Frictional Force

The maximum static frictional force that can act on the girl while she runs is given by:

• F_friction = µ * N

Here, N is the normal force, which equals the weight of the girl:

• N = m_g * g = 50 \, \text{kg} * 10 \, \text{m/s}^2 = 500 \, \text{N}

Now, substituting the values:

• F_friction = (1/3) * 500 \, \text{N} = 166.67 \, \text{N}

Applying Newton's Second Law

When the girl runs to the right, she exerts a force on the trolley to the left due to Newton's third law. This frictional force will cause the trolley to accelerate to the left. We can set up the equations of motion for both the girl and the trolley.

Acceleration of the Girl

The acceleration of the girl can be calculated using Newton's second law:

• F = m * a

Thus, the acceleration of the girl (a_g) is:

• a_g = F_friction / m_g = 166.67 \, \text{N} / 50 \, \text{kg} = 3.33 \, \text{m/s}^2

Acceleration of the Trolley

Now, the trolley will also accelerate due to the frictional force acting on it:

• F = m * a

The acceleration of the trolley (a_t) is:

• a_t = F_friction / m_t = 166.67 \, \text{N} / 100 \, \text{kg} = 1.67 \, \text{m/s}^2

Relative Motion Consideration

Since the girl runs with a maximum speed of 9 m/s relative to the trolley, we need to consider the relative motion between the girl and the trolley. The total acceleration of the girl with respect to the ground can be expressed as:

• a_relative = a_g + a_t

Substituting the values:

• a_relative = 3.33 \, \text{m/s}^2 + 1.67 \, \text{m/s}^2 = 5 \, \text{m/s}^2

Finding the Time to Reach Maximum Speed

To find the time (t) it takes for the girl to reach her maximum speed of 9 m/s, we can use the formula:

• v = u + at

Where:

• v = final velocity (9 m/s)
• u = initial velocity (0 m/s)
• a = acceleration (3.33 m/s²)

Rearranging the equation to solve for time:

• t = (v - u) / a = (9 \, \text{m/s} - 0) / 3.33 \, \text{m/s}^2 ≈ 2.7 \, \text{s}

Final Calculation

Thus, the minimum time in which the girl can acquire her maximum speed, without slipping, is approximately:

• t ≈ 2.7 seconds

This analysis shows how the forces and accelerations interact in a system where friction plays a crucial role in the motion of both the girl and the trolley. Understanding these dynamics is essential in physics, especially in mechanics.