AP Physics C Mechanics: 2.3 Newton’s Laws of Motion: Third Law – Exam Style questions with Answer- FRQ

Question

A mass \(\mathrm{m}_2=1.0 \mathrm{~kg}\), on a \(36.1^{\circ}\) incline, is connected to a mass \(\mathrm{m}_1=6.1 \mathrm{~kg}\), on a horizontal surface. The surfaces and the pulley are frictionless. If \(F=22.5 \mathrm{~N}\), what is the magnitude of the tension in the connecting cord?

▶️Answer/Explanation

Ans:

Applying $F=ma$ for whole system ,

$F-m_2gsin\theta=(m_1+m_2)a_{sys}$

$22.5-1\times 9.8\times sin(36.1^{\circ})=(1+6.1)a_{sys}$

$22.5-5.77=7.1\times a_{sys}$

$a_{sys}=\frac{16.72}{7.1} \Rightarrow 2.355~\rm{m/s^2}$

Now , applying $F=ma$ for $m_1$

$T=m_1(a_{sys})$

$T=6.\times 2.355 \Rightarrow 14.3~\rm{N}$

 

Question

An object of mass m moving along the x-axis with velocity v is slowed by a force F = -kv, where k is a constant. At time t = 0, the object has velocity vo at position x = 0, as shown above.
a. What is the initial acceleration (magnitude and direction) produced by the resistance force?
b. Derive an equation for the object’s velocity as a function of time t, and sketch this function on the axes below. Let a velocity directed to the right be considered positive.

c. Derive an equation for the distance the object travels as a function of time t and sketch this function on the axes below.

d. Determine the distance the object travels from t = 0 to t = ∞.

Answer/Explanation

Ans:

a. F = ma; F = –kv = ma; a0 = –kv0
b.

c.

d. at t = ∝, 𝑥 = \(\frac{mv_{0}}{k}\)

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