Laws of Motion

JEE Main level, numerical answer, medium

131 of 210

Question 131 of 210

In the given figure the blocks A,BA, B and CC weigh 4 kg,6 kg4 \mathrm{~kg}, 6 \mathrm{~kg} and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force F⃗\vec{F} required to slide the block CC with constant speed is ____\_\_\_\_ N.
(Use g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^{2})

Show answer and solution

Answer: 210

For the 8 kg block (C) to move with constant velocity, Fnet=0F_{net} = 0.

Friction forces acting:
- Between C and ground: f=μ(mA+mB+mC)g=0.5×(4+6+8)×10=90f = \mu(m_A + m_B + m_C)g = 0.5 \times (4+6+8) \times 10 = 90 N
- Between C and B: f=μ(mA+mB)g=0.5×(4+6)×10=50f = \mu(m_A + m_B)g = 0.5 \times (4+6) \times 10 = 50 N

For the 6 kg block (B):
T=fB on ground+fA on BT = f_{\text{B on ground}} + f_{\text{A on B}}
T=μ(mA+mB)g⋅mBmA+mB+μ⋅mA⋅gT = \mu(m_A + m_B)g \cdot \frac{m_B}{m_A+m_B} + \mu \cdot m_A \cdot g
T=20+50=70T = 20 + 50 = 70 N

For the 8 kg block (C):
F=90+T+50=90+70+50=210F = 90 + T + 50 = 90 + 70 + 50 = 210 N

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Laws of motion, friction and pulleys simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Laws of Motion (Frictional force), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.