Question

give me some solved questions on block on block friction

Answer 1

Here's a solved question on block-on-block friction, suitable for JEE/NEET preparation.

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Question

A block A of mass 2 kg is placed on a larger block B of mass 4 kg. The coefficient of static friction between block A and block B is 0.4. Block B rests on a smooth horizontal surface (i.e., no friction between block B and the ground). A horizontal force $F$ is applied to block A. (Take $g = 10 \text{ m/s}^2$).

(a) Find the maximum force $F$ that can be applied to block A such that both blocks move together without slipping. (b) If $F = 20$ N, find the accelerations of block A and block B.

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Visual Representation

Free Body Diagrams (FBDs)

Block A (mass $m_1 = 2 \text{ kg}$):

• Vertical forces:
  • Weight: $m_1 g$ (downwards)
  • Normal force from B: $N_1$ (upwards)
• Horizontal forces:
  • Applied force: $F$ (to the right)
  • Friction from B: $f$ (to the left, opposing relative motion of A w.r.t B)

Block B (mass $m_2 = 4 \text{ kg}$):

• Vertical forces:
  • Weight: $m_2 g$ (downwards)
  • Normal force from A: $N_1$ (downwards, action-reaction pair with $N_1$ on A)
  • Normal force from ground: $N_2$ (upwards)
• Horizontal forces:
  • Friction from A: $f$ (to the right, action-reaction pair with $f$ on A)
  • No friction from the ground (smooth surface).

Part (a): Maximum force $F$ for no slipping

When both blocks move together, they have a common acceleration, let's call it $a$.

1. Vertical equilibrium for Block A: $N_1 - m_1 g = 0 \implies N_1 = m_1 g = 2 \text{ kg} \times 10 \text{ m/s}^2 = 20 \text{ N}$

2. Maximum static friction between A and B: $f_{s,max} = \mu_s N_1 = 0.4 \times 20 \text{ N} = 8 \text{ N}$

3. Horizontal motion for Block B: The only horizontal force on Block B is the friction force $f$ from Block A. This force causes Block B to accelerate. $f = m_2 a$ For no slipping, the friction force $f$ required for common acceleration must not exceed $f_{s,max}$: $f \le f_{s,max} \implies m_2 a \le 8 \text{ N}$ $4 \text{ kg} \times a \le 8 \text{ N} \implies a \le 2 \text{ m/s}^2$ So, the maximum common acceleration is $a_{max} = 2 \text{ m/s}^2$.

4. Horizontal motion for the combined system (A+B): When moving together, we can treat (A+B) as a single system of mass $(m_1 + m_2)$. The external horizontal force is $F$. $F = (m_1 + m_2) a$ To find the maximum force $F_{max}$ for no slipping, we use $a_{max}$: $F_{max} = (2 \text{ kg} + 4 \text{ kg}) \times 2 \text{ m/s}^2 = 6 \text{ kg} \times 2 \text{ m/s}^2 = 12 \text{ N}$

Part (b): Accelerations if $F = 20$ N

Since the applied force $F = 20 \text{ N}$ is greater than $F_{max} = 12 \text{ N}$, the blocks will slip relative to each other. In this case, the friction between A and B becomes kinetic friction. We assume $\mu_k = \mu_s = 0.4$ unless specified otherwise.

1. Kinetic friction force between A and B: $f_k = \mu_k N_1 = 0.4 \times 20 \text{ N} = 8 \text{ N}$

2. Acceleration of Block A ($a_A$): Applying Newton's second law to Block A: $F - f_k = m_1 a_A$ $20 \text{ N} - 8 \text{ N} = 2 \text{ kg} \times a_A$ $12 \text{ N} = 2 \text{ kg} \times a_A \implies a_A = \frac{12}{2} = 6 \text{ m/s}^2$

3. Acceleration of Block B ($a_B$): Applying Newton's second law to Block B: $f_k = m_2 a_B$ $8 \text{ N} = 4 \text{ kg} \times a_B \implies a_B = \frac{8}{4} = 2 \text{ m/s}^2$

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Explanation of the Solution

(a) For no slipping, both blocks move with the same acceleration $a$. The friction force $f$ between A and B is responsible for accelerating block B. Thus, $f = m_2 a$. This friction force cannot exceed the maximum static friction, $f_{s,max} = \mu_s m_1 g$. So, $m_2 a \le \mu_s m_1 g$, which gives the maximum common acceleration $a_{max}$. The total external force $F$ required to accelerate the combined system $(m_1+m_2)$ with $a_{max}$ is $F_{max} = (m_1+m_2)a_{max}$.

(b) If $F > F_{max}$, the blocks slip. The friction between them becomes kinetic friction, $f_k = \mu_k m_1 g$. Now, apply Newton's second law separately to each block using this kinetic friction: for block A, $F - f_k = m_1 a_A$; for block B, $f_k = m_2 a_B$. Solve for $a_A$ and $a_B$.

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Answer

(a) The maximum force $F$ that can be applied such that both blocks move together without slipping is 12 N. (b) If $F = 20$ N, the acceleration of block A is $6 \text{ m/s}^2$ and the acceleration of block B is $2 \text{ m/s}^2$.

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Answer 2

Explanation of the solution is provided in the correct_answer