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AP Physics

Unit 2 - Linear Forces

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Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle θ=32∘ \theta = 32^\circ above the horizontal. If the coefficients of friction are μA=0.2 \mu_A = 0.2 and μB=0.3 \mu_B = 0.3 and if mA=mB=5.0 m_A = m_B = 5.0 kg \text{kg} , determine:

  1. (a) The acceleration of the blocks. 3 points

  2. (b) The tension in the cord between the blocks. 3 points

Verfied Answer
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Part (a): Acceleration of the blocks

Step Derivation/Formula Reasoning
1 Fgx=mgsin⁡θ F_{g_x} = m g \sin \theta Calculate the component of gravitational force along the incline for one block.
2 FfA=μAmAgcos⁡θ F_{f_A} = \mu_A m_A g \cos \theta Calculate the frictional force for block A. Frictional force is the product of the friction coefficient, mass, gravitational acceleration, and the cosine of the incline angle.
3 FfB=μBmBgcos⁡θF_{f_B} = \mu_B m_B g \cos \theta Calculate the frictional force for block B using its coefficient of friction.
4 Fnet=2Fgx–FfA–FfB F_{\text{net}} = 2 F_{g_x} – F_{f_A} – F_{f_B} Net force is the sum of both gravitational components minus the frictional forces for both blocks.
5 Fnet=2(mgsin⁡θ)–(μAmgcos⁡θ F_{\text{net}} = 2 (m g \sin \theta) – (\mu_A m g \cos \theta
+μBmgcos⁡θ)+ \mu_B m g \cos \theta) 		Substitute the expressions for gravitational and frictional forces into the net force equation.
6 Fnet= 2⋅5⋅9.8sin⁡(32)–(0.2⋅5⋅9.8cos⁡(32) F_{\text{net}} = 2 \cdot 5 \cdot 9.8 \sin(32) – (0.2 \cdot 5 \cdot 9.8 \cos(32)

+0.3⋅5⋅9.8cos⁡(32))+ 0.3 \cdot 5 \cdot 9.8 \cos(32))

Substitute known values: m=5 kg, g=9.8 m/s2, θ=32∘ m = 5 \ \text{kg}, \ g = 9.8 \ \text{m/s}^2, \ \theta = 32^\circ .
7 Fnet=52.012–8.3–12.5= F_{\text{net}} = 52.012 – 8.3 – 12.5 =

31.2 N31.2 \ \text{N}

Calculate the net force acting on both blocks by subtracting the calculated frictional forces from the gravitational force component.
8 a=Fnet2m=31.22×5a = \frac{F_{\text{net}}}{2m} = \frac{31.2}{2 \times 5} Use Newton’s second law F=ma F = ma to solve for acceleration.
9 3.1 m/s2 \boxed{3.1 \ \text{m/s}^2} Calculate the acceleration of the blocks.

Part (b): Tension in the cord

Step Derivation/Formula Reasoning
1 T=mAgsin⁡θ–FfA–mAa T = m_A g \sin \theta – F_{f_A} – m_A a Consider block A. The tension is the net force after subtracting the force needed for acceleration and friction.
2 T=5⋅9.8sin⁡(32)–(0.2⋅5⋅9.8cos⁡(32))–5⋅3.1 T = 5 \cdot 9.8 \sin(32) – (0.2 \cdot 5 \cdot 9.8 \cos(32)) – 5 \cdot 3.1 Substitute known values for mass, gravitational acceleration, incline angle, friction coefficient, and acceleration.
3 T=26–8.3–15.6=2.1 NT = 26 – 8.3 – 15.6 = 2.1 \ \text{N} Calculate the tension in the cord between the blocks.
4 2.1 N \boxed{2.1 \ \text{N}} Express the tension in the cord as the final answer.

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  1. 3.1 m/s23.1 \, \text{m/s}^2
  2. 2.1 N2.1 \, \text{N}

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