Question

Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in m/s² is : (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless) :

• A. 3
• B. 4
• C. 2
• D. 1.2

Answer 1

The Correct Option is C

The kinetic frictional force is given by:

\[ f_k = \mu N = 0.04 \times 20g = 8 \, \text{Newton} \]

The net force acting on the system is:

\[ F_{\text{net}} = 60 \, \text{N} - f_k = 60 - 8 = 52 \, \text{N} \]

The total mass of the system is:

\[ m_{\text{total}} = 20 \, \text{kg} + 6 \, \text{kg} = 26 \, \text{kg} \]

The acceleration of the system is:

\[ a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{52}{26} = 2 \, \text{ms}^{-2} \]

Answer 2

The problem asks for the acceleration of a system consisting of a 20 kg trolley on a frictional horizontal surface, connected by a string over a frictionless pulley to a 6 kg hanging block.

Concept Used:

1. Newton's Second Law of Motion: The net force acting on an object or a system is equal to the product of its mass and acceleration (\( F_{\text{net}} = ma \)). We will apply this law to the system as a whole.

2. Force of Kinetic Friction: When an object moves on a surface, the force of kinetic friction (\(f_k\)) opposes its motion. It is calculated as:

\[ f_k = \mu_k N \]

where \(\mu_k\) is the coefficient of kinetic friction and \(N\) is the normal force. For an object on a horizontal surface, the normal force is equal to its weight (\(N = mg\)).

3. System Analysis: For a connected system, the net driving force is the gravitational force on the hanging mass, and the opposing force is the friction on the trolley. The total mass being accelerated is the sum of the masses of both objects.

\[ a = \frac{F_{\text{driving}} - F_{\text{opposing}}}{M_{\text{total}}} \]

Step-by-Step Solution:

Step 1: List the given parameters.

Mass of the trolley, \(m_1 = 20 \, \text{kg}\).

Mass of the hanging block, \(m_2 = 6 \, \text{kg}\).

Weight of the hanging block, \(W_2 = 60 \, \text{N}\). From this, we can infer the acceleration due to gravity, \(g = \frac{W_2}{m_2} = \frac{60 \, \text{N}}{6 \, \text{kg}} = 10 \, \text{m/s}^2\).

Coefficient of kinetic friction, \(\mu_k = 0.04\).

Step 2: Calculate the force of kinetic friction (\(f_k\)) on the trolley.

First, find the normal force (\(N\)) acting on the trolley. Since the trolley is on a horizontal surface, the normal force is equal to its weight.

\[ N = m_1 g = (20 \, \text{kg}) \times (10 \, \text{m/s}^2) = 200 \, \text{N} \]

Now, calculate the kinetic friction force using its formula:

\[ f_k = \mu_k N = 0.04 \times 200 \, \text{N} = 8 \, \text{N} \]

Step 3: Determine the net force acting on the system.

The driving force for the system is the weight of the hanging block, \(W_2 = 60 \, \text{N}\).

The opposing force is the kinetic friction on the trolley, \(f_k = 8 \, \text{N}\).

The net force (\(F_{\text{net}}\)) causing the acceleration of the system is:

\[ F_{\text{net}} = W_2 - f_k = 60 \, \text{N} - 8 \, \text{N} = 52 \, \text{N} \]

Step 4: Calculate the total mass of the system.

The total mass (\(M_{\text{total}}\)) that is being accelerated is the sum of the masses of the trolley and the hanging block.

\[ M_{\text{total}} = m_1 + m_2 = 20 \, \text{kg} + 6 \, \text{kg} = 26 \, \text{kg} \]

Step 5: Calculate the acceleration of the system using Newton's Second Law.

Using the formula \(F_{\text{net}} = M_{\text{total}} \cdot a\), we can solve for the acceleration \(a\).

\[ a = \frac{F_{\text{net}}}{M_{\text{total}}} \]

Substitute the values calculated in the previous steps:

\[ a = \frac{52 \, \text{N}}{26 \, \text{kg}} \] \[ a = 2 \, \text{m/s}^2 \]

The acceleration of the system is 2 m/s².