Question

Two blocks A and B of masses 2 kg and 4 kg respectively are kept on a rough horizontal surface. If same force of 20 N is applied on each block, then the ratio of the accelerations of the blocks A and B is (Coefficient of kinetic friction between the surface and the blocks is 0.3 and acceleration due to gravity = \(10 \operatorname{ms}^{-2}\))

• A. \(1:1 \)
• B. \(7:2 \)
• C. \(1:2 \)
• D. \(4:3 \)

Hint:

When dealing with forces and acceleration on a rough surface: 1. Draw a Free Body Diagram (FBD): Identify all forces acting on the object (applied force, friction, normal force, gravity). 2. Calculate Normal Force (N): On a horizontal surface, \(N = mg\). 3. Calculate Friction Force (\(f_k\) or \(f_s\)): Kinetic friction \(f_k = \mu_k N\). Maximum static friction \(f_{s,max} = \mu_s N\). 4. Apply Newton's Second Law (\(F_{net} = ma\)): Sum forces in the direction of motion. If the applied force exceeds friction, the object accelerates.

Answer 1

The Correct Option is B

Step 1: Identify the given parameters and general formula for acceleration.
Given:
Mass of block A, \(m_A = 2 \operatorname{kg}\)
Mass of block B, \(m_B = 4 \operatorname{kg}\)
Applied force on each block, \(F = 20 \operatorname{N}\)
Coefficient of kinetic friction, \(\mu_k = 0.3\)
Acceleration due to gravity, \(g = 10 \operatorname{ms}^{-2}\)
For a block on a rough horizontal surface with an applied force \(F\), the net force acting on the block in the direction of motion is \(F_{net} = F - f_k\), where \(f_k\) is the kinetic friction force. The kinetic friction force is given by \(f_k = \mu_k N\), where \(N\) is the normal force. For a horizontal surface, \(N = mg\). So, \(f_k = \mu_k mg\). According to Newton's second law, \(F_{net} = ma\), so \(F - \mu_k mg = ma\).
Thus, the acceleration \(a\) is given by: \[ a = \frac{F - \mu_k mg}{m} \] Before calculating, we must ensure that the applied force is greater than the maximum static friction (or in this case, kinetic friction is assumed to be active, implying motion). The kinetic friction force must be less than the applied force for acceleration to occur. 
Step 2: Calculate the acceleration for block A (\(a_A\)).
For block A:
Friction force \(f_{k,A} = \mu_k m_A g = (0.3)(2 \operatorname{kg})(10 \operatorname{ms}^{-2}) = 6 \operatorname{N}\).
Since \(F = 20 \operatorname{N}>f_{k,A} = 6 \operatorname{N}\), the block accelerates. \[ a_A = \frac{F - f_{k,A}}{m_A} = \frac{20 \operatorname{N} - 6 \operatorname{N}}{2 \operatorname{kg}} = \frac{14 \operatorname{N}}{2 \operatorname{kg}} = 7 \operatorname{ms}^{-2} \] Step 3: Calculate the acceleration for block B (\(a_B\)).
For block B:
Friction force \(f_{k,B} = \mu_k m_B g = (0.3)(4 \operatorname{kg})(10 \operatorname{ms}^{-2}) = 12 \operatorname{N}\).
Since \(F = 20 \operatorname{N}>f_{k,B} = 12 \operatorname{N}\), the block accelerates. \[ a_B = \frac{F - f_{k,B}}{m_B} = \frac{20 \operatorname{N} - 12 \operatorname{N}}{4 \operatorname{kg}} = \frac{8 \operatorname{N}}{4 \operatorname{kg}} = 2 \operatorname{ms}^{-2} \] Step 4: Determine the ratio of the accelerations.
The ratio of the accelerations of blocks A and B is \(a_A : a_B\): \[ \frac{a_A}{a_B} = \frac{7 \operatorname{ms}^{-2}}{2 \operatorname{ms}^{-2}} = \frac{7}{2} \] So the ratio is 7:2. The final answer is $\boxed{7:2}$.