University

Physics 1

Physics 1

Fall 2024

Blocks \(A\) and \(B\) are connected by a string passing over a pulley. Block \(B\) if moving down, dragging block \(A\) across a frictionless table.

- Sketch the forces on blocks \(A\) and \(B\).
- Considering the axes shown, state the acceleration constraint for this system. (That is, what equation relates the accerlations of \(A\) and \(B\)?)

- \(a_{Ax} = -a_{By}\)

A hockey puck is sliding frictionlessly in the \(+x\) direction. When it reaches the position marked by the dot, you must give it one sharp blow with a hammer. After hitting it you want the puck to move in the \(-y\) direction at a speed similar to what it had initially.

Draw a force vector to show the direction in which you exert the force of the hammer.

The motion diagram shows a dot at the location of an object every second. Draw the net force vectors at each point.

Blocks \(A\) and \(B\), with \(m_B > m_A\), are connected by a string. A hand pushes \(A\) and accelerates the system across a frictionless surface.

Rank the magnitudes of the following forces:

- \(T_A\), the string tension force acting on \(A\)
- \(T_B\), the string tension force acting on \(B\)
- \(H_A\), the force of the hand on \(A\)

Since the masses are accelerating to the right, both free-body diagrams should have the net rightward force greater than the net leftward force.

- \(H_A > T_A\) to accelerate \(A\) to the right
- \(T_A = T_B\) because tension is constant throughout the string

So:

\(H_A > T_A = T_B\)

Blocks \(A\) and \(B\) are connected by a massless string over an ideal pulley. The blocks have just been released from rest. Rank the following forces: the weight of \(A\) (\(W_A\)), the weight of \(B\) (\(W_B\)) and the tension in the string (\(T\))?

Block \(A\) must have net force downward. Block \(B\) must have net force upward.

\(W_A > T > W_B\)

A woman stands on a scale in an elevator. The velocity of each is shown, along with its change in speed. Rank the scale readings, \(S_1\), \(S_2\) and \(S_3\).

\(S_1 = S_2 \gt S_3\)

The only two forces on the person are gravity downward (which does not change) and scale force \(S\) upward. If \(\vec{a}\) points upward, then the net force points upward. This is true for case 1 and 2.

A history book is lying on top of a physics book on a desk, as shown. The history and physics books weigh 14 N and 18 N, respectively.

Sketch a free-body diagram (FBD) for each book. Identify each force with a double subscript notation (for instance, the contact force by the history book on the physics book can be described as \(\vec{F}_{hp}\)).

Determine the magnitude of each force in the FBDs.

Force magnitudes are as follows

\(F_{eh} = 14\) N. This is the force by the Earth on the history book.

\(F_{ph} = 14\) N. This must balance \(F_{eh}\)

\(F_{ep} = 18\) N. This is the force by the Earth on the physics book.

\(F_{hp} = 14\) N. This is the 3rd Law pair to \(F_{ph}\).

\(F_{dp} = 32\) N. This is the sum of \(F_{hp} + F_{ep}\).

Two forces are applied to a \(5.0\) kg object, and it accelerates at a rate of \(2.0\) m/s² in the positive \(y\)-direction. If one of the forces acts in the positive \(x\)-direction with magnitude \(12.0\) N, find the magnitude of the other force.

\(F_2 = 15.6\) N

\(\displaystyle \vec{F}_{\rm{net}} = m \vec{a}\)

\((12.0) \hat{\imath} + \vec{F}_2 = (5.0) (2.0) \hat{\jmath}\)

\(\vec{F}_2 = -12 \hat{\imath} +10 \hat{\jmath}\)

\(F_2 = \sqrt{(-12)^2 + 10^2} = 15.6\) N

Two identical springs, each with the spring constant \(20\) N/m, support a \(15.0\) N weight.

- What is the tension in spring A?
- What is the amount of stretch of spring A from its equilibrium position?

- \(8.66\) N
- \(0.433\) m

Procedure:

- Draw a free body diagram for the weight.
- This will have two identical spring forces and one weight force.
- The three vertical forces add to zero, which yields an equation for the spring tension.
- With this tension, use \(F=kx\) to find the spring extension \(x\).

A force acts on a car of mass \(m\) so that the position of the car changes in time according to \(\vec{r}(t)= (At) \hat{\imath} + (B/t) \hat{\jmath}\).

Find the force vector acting on the car as a function of time.

\[ \vec{F} = m \vec{a} = m \frac{d}{dt} \frac{d\vec{r}}{dt} = m \frac{d}{dt} \left ( A \hat{\imath} - \frac{B}{t^2} \hat{\jmath} \right ) = \frac{2mB}{t^3} \hat{\jmath} \]

For the connected masses shown, \(M = 6.0\) kg and \(\theta = 30^{\circ}\). Find the acceleration of the system and the tension in the connecting string. The pulley and surfaces are frictionless.

\(a = 7.35\) m/s²

\(T=14.7\) N

Start with a free-body diagram for each mass. It’s ok if each FBD has
its own coordinate system; what’s important is that the scalar variables
(\(T\), \(a\), \(M\)) used in multiple figures all refer to
the *same* quantity.

Next we write \(\vec{F}_{\rm{net}} = m \vec{a}\) in each case.

For the mass on the incline:

\(T \hat{\imath} + N \hat{\jmath} + Mg ( \hat{\imath} \sin \theta - \hat{\jmath} \cos \theta ) = Ma \hat{\imath}\)

which is really two equations:

\(T + Mg \sin \theta = Ma\)

\(N - Mg \cos \theta = 0\)

For the hangning mass:

\(T \hat{\jmath} -Mg \hat{\jmath} = M(-a \hat{\jmath})\)

or

\(T -Mg = -Ma\)

The equations that contain \(T\) and \(a\) must be solved simultaneously. One way to do this is to take their ratio:

\(\displaystyle \frac{ T + Mg \sin \theta}{T-Mg} = -1\)

\(\displaystyle T = \tfrac{1}{2} Mg ( 1- \sin \theta) =14.7\) N

Plug this result into the hanging mass equation to find

\(a = 7.35\) m/s².

*Last modified: Tue October 22 2024, 10:11 PM.*