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SAT II Physics : Linear Motion

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Example Question #1 : Kinematics Equations

If the block of mass $\displaystyle m$ slides up a frictionless incline plane and is pulled by mass $\displaystyle M$ which is falling then what is the acceleration of the block on the ramp if m= 20kg $\displaystyle m= 20kg$ and M=40kg $\displaystyle M=40kg$ and the angle of incline is 30 degrees?

Possible Answers:

a=2g $\displaystyle a=2g$

$a=g\sqrt{2}$ $\displaystyle a=g\sqrt{2}$

$a=\frac{1}{2}g$ $\displaystyle a=\frac{1}{2}g$

$a=\frac{\sqrt{3}}{2}g$ $\displaystyle a=\frac{\sqrt{3}}{2}g$

$a=\frac{5}{6}g$ $\displaystyle a=\frac{5}{6}g$

Correct answer:

$a=\frac{1}{2}g$ $\displaystyle a=\frac{1}{2}g$

Explanation:

Begin by making a free body diagram for each block:

Q62

Use the diagram to write an equation for net force on each block:

$F_{net} =F_{t}-mg \sin (30)$ $\displaystyle F_{net} =F_{t}-mg \sin (30)$

$F_{net} = Mg -F_{t}$ $\displaystyle F_{net} = Mg -F_{t}$

Since F=ma $\displaystyle F=ma$ for the block on the ramp and F=Ma $\displaystyle F=Ma$ for falling block we can substitute into these equations:

$ma =F_{t}-mg \sin (30)$ $\displaystyle ma =F_{t}-mg \sin (30)$

$Ma = Mg -F_{t}$ $\displaystyle Ma = Mg -F_{t}$

Then add the equations to get:

$Ma+ma = Mg-mg\sin (30)$ $\displaystyle Ma+ma = Mg-mg\sin (30)$

Next rearrange to isolate $\displaystyle a$:

$a = \frac{M-msin(30)}{M+m}$ $\displaystyle a = \frac{M-msin(30)}{M+m}$

Substitute the given values from the question and solve:

$a = \frac{40-20sin(30)}{40+20}$ $\displaystyle a = \frac{40-20sin(30)}{40+20}$

$a=\frac{1}{2}g$ $\displaystyle a=\frac{1}{2}g$

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SAT II Physics : Linear Motion

Study concepts, example questions & explanations for SAT II Physics

All SAT II Physics Resources

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Example Question #1 : Kinematics Equations

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