Kinetic and Static Friction

AP Physics 1· difficulty 4/5

Two blocks A (3 kg) and B (5 kg) sit in contact on a floor with μk=0.20\mu_k = 0.20. A horizontal force F=40 NF = 40\text{ N} pushes A from the left. What is the contact force between A and B? Use g=9.8 m/s2g = 9.8\text{ m/s}^2.

  • A

    18.4 N

  • B

    25.0 N

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  • C

    30.0 N

  • D

    15.0 N

Explanation

System acceleration: a=(Fμk(mA+mB)g)/(mA+mB)=(400.20×8×9.8)/8=(4015.68)/83.04 m/s2a = (F - \mu_k(m_A+m_B)g)/(m_A+m_B) = (40 - 0.20 \times 8 \times 9.8)/8 = (40 - 15.68)/8 \approx 3.04\text{ m/s}^2. Force on B alone: FAB=mB(a+μkg)=5(3.04+1.96)=5×5=25.0 NF_{AB} = m_B(a + \mu_k g) = 5(3.04 + 1.96) = 5 \times 5 = 25.0\text{ N}.

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