Newton’s Second Law Diagrams AP Physics C Mechanics MCQ – Exam Style Questions etc.
Newton’s Second Law Diagrams AP Physics C Mechanics MCQ
Unit 2: Force and Translational Dynamics
Weightage : 20-15%
Question
The momentum p of a moving object as a function of time t is given by the expression \(p = kt^3\), where k is a constant. The force causing this motion is given by the expression
(A) \(3kt^2\)
(B)\(\frac{3kt^2}{2}\)
(C)\(\frac{kt^2}{2}\)
(D)\( kt^4\)
(E) \(\frac{kt^4}{4}\)
Answer/Explanation
Ans:A
An object of mass 0.5 kg is given an initial velocity and then slides across a horizontal surface. The object experiences a resistive force that is a function of velocity. The velocity v of the object as a function of time t is given by \(v(t)=\alpha e^{-\beta t}\)=,where\( \alpha =2m/s\) and\( \beta =3s^{-1}\).
Question
Which of the following is a correct expression for the net force, in newtons, exerted on the object as a function of time?
(A)\(6e^{-3t}\)
(B)\(-3e^{-3t}\)
(C)\(e^{-3t}\)
(D)\(2e^{-3t}\)
(E)\(\frac{2}{3\left ( 1-e^{-3t} \right )}\)
Answer/Explanation
Ans:B
The acceleration is the derivative of the velocity, \(a=\frac{dv}{dt}=\frac{dv}{t}(\alpha e^{\beta t})=-\beta \alpha e^{-\beta t}\)Substituting into an equation for
Newton’s second law yields
\(F_{net}=ma=m-\beta \alpha e^{-\beta t}\)
\(F_{net}=(0.5kg)(-3s^{-1})(2m/s)e^(-3s^{-1})=-3e^{-3t}\)
Question
A block of mass 3 kg, initially at rest, is pulled along a frictionless, horizontal surface with a force shown as a function of time t by the graph above.
The acceleration of the block at t = 2 s is
(A) \(3/4 m/s^2\)
(B) \(4/3 m/s^2\)
(C) \(2 m/s^2\)
(D) \(8 m/s^2\)
(E) \(12 m/s^2\)
Answer/Explanation
Ans:B
Question
A block of mass 3 kg, initially at rest, is pulled along a frictionless, horizontal surface with a force shown as a function of time t by the graph above.
The speed of the block at t = 2 s is
(A) 4/3 m/s
(B) 8/3 m/s
(C) 4 m/s
(D) 8 m/s
(E) 24 m/s
Answer/Explanation
Ans:A