Kinetic and Static Friction

AP Physics 1· difficulty 4/5

m₁ m₂ F

A 2.0 kg block sits atop a 5.0 kg block on a frictionless floor. Static friction between blocks is μs=0.40\mu_s = 0.40. What is the maximum horizontal force FF on the bottom block such that the top block does not slide? Use g=9.8 m/s2g = 9.8\text{ m/s}^2.

  • A

    19.6 N

  • B

    39.2 N

  • C

    13.7 N

  • D

    27.4 N

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Explanation

Top block accelerates only via friction: amax=μsg=3.92 m/s2a_{\max} = \mu_s g = 3.92\text{ m/s}^2. The system has total mass 7.0 kg, so F=(m1+m2)amax=7.0×3.92=27.4 NF = (m_1 + m_2) a_{\max} = 7.0 \times 3.92 = 27.4\text{ N}.

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