University

Physics 1

Physics 1

Fall 2024

A boy is pushing a box across the floor toward the right at a steadily increasing speed. There is some friction. Sketch the vectors for a free-body diagram for the box.

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

Rank the magnitudes of the following forces:

- \(T_A\), the string tension force acting on \(A\)
- \(T_B\), the string tension force acting on \(B\)
- \(f_A\), the friction force acting on \(A\)
- \(f_B\), the friction 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.

- \(H_A > T_A +f_A\) to accelerate \(A\) to the right
- \(T_A = T_B\) because tension is constant throughout the string
- \(T_B > f_B\) to accelerate \(B\) to the right
- \(f_B > f_A\) because \(B\) has greater mass than \(A\)

So:

\(H_A > T_A = T_B > f_B > f_A\)

A car is parked on a hill.

- Draw the free-body diagram for the car, identifying the forces.
- What is the direction of the net force on the car?

- \(F_G\) is gravity, \(N\) is the normal force, \(f_s\) is static friction

- Net force is zero since it is stationary.

In a game of tetherball, a ball on a string swings around in a horizontal circle at a constant speed. At the moment shown, what are the directions of

- the tension force on the ball
- the gravity force on the ball
- the net force on the ball

- up to the right, parallel to the string
- straight down
- horizontal to the right (the net force is purely centripetal)

In case a, block \(A\) is accelerated across a frictionless table by hanging a 10 N weight. In case b, the same block is accelerated by a steady 10 N tension in the string. State if case a or b has the greater…

- tension
- acceleration

- Tension is greater in case b. In case a the tension must be less than \(10\) N since the hanging weight accelerates downward.
- Acceleration is greater in case b. This is due to the higher tension.

Suppose you press a book (toward the right) against a wall with your hand. The book is not moving.

- Draw a free-body diagram of the book, identifying the forces.
- Now suppose you decrease the strength of your push, but not enough
for the book to slip. What happens to the magnitude of each force? (Use
*increase*,*decrease*, or*remain the same*.)

- \(F_{\rm{push}}\) is force from
hand, \(F_G\) is gravity, \(N\) is normal force, \(f_s\) is static friction

- \(F_{\rm{push}}\) decreases, \(F_G\) remains the same, \(N\) decreases, \(f_s\) remains the same

Consider a crate with mass \(m\) on the back of a flatbed truck. There are no straps or any retaining mechanisms to keep the crate in place. The static and kinetic friction coefficients between the flatbed and the crate are \(\mu_s\). and \(\mu_k\), respectively.

- At first the truck accelerates very slowly, and the crate moves with the truck without slipping.
- Then the truck increases its acceleration, and the crate is about to slip off the truck.
- Then the truck increases its acceleration further, and the crate slides toward the back of the truck.

- During motion 1, draw a free-body diagram of the crate, identifying the forces.
- During motion 1, is the friction force greater than, less than or equal to \(\mu_s mg\)?
- During motion 3, draw a free-body diagram of the crate, identifying the forces.
- During motion 3, is the friction force greater than, less than or equal to \(\mu_k mg\)?

- \(F_G\) is gravity, \(N\) is normal force, \(f_s\) is static friction

- Static friction is less than \(\mu_s mg\), which is the friction at the point of impending slip.
- \(F_G\) is gravity, \(N\) is normal force, \(f_k\) is kinetic friction

- Kinetic friction, when present, is always equal to \(\mu_k mg\).

Located at the origin, an electric car of mass \(m\) is at rest and in equilibrium. A time dependent force of \(\vec F(t)\) is applied at time \(t=0\) , and its components are \(F_x(t)=p+nt\) and \(F_y(t)=qt\) where \(p\), \(q\), and \(n\) are constants. Find the position \(\vec r(t)\) and velocity \(\vec v(t)\) as functions of time \(t\).

\(\displaystyle \vec v(t) = \left ( \frac{pt}{m} + \frac{nt^2}{2m} \right ) \hat{\imath} + \left ( \frac{qt^2}{2m} \right ) \hat{\jmath}\)

\(\displaystyle \vec r(t) = \left ( \frac{pt^2}{2m} + \frac{nt^3}{6m} \right ) \hat{\imath} + \left ( \frac{qt^3}{6m} \right ) \hat{\jmath}\)

The mass of block 1 is \(m_1\) = \(4.0\) kg, while the mass of block 2 is \(m_2\) = \(8.0\) kg. The coefficient of friction between \(m_1\) and the inclined surface is \(\mu_k\) = \(0.40\). After the masses are released from rest, they begin to move.

Find the acceleration of the system and the tension in the string.

\(F_{net} = m_2 g - m_1 g \sin \theta - \mu m_1 g \cos \theta = (m_1 +m_2) a\)

\(a = 3.52\) m/s²

For mass 2,

\(T-m_2 g = -m_2 a\), so

\(T = 50.2\) N

A crate having mass 50.0 kg falls horizontally off the back of the flatbed truck, which is traveling at 100 km/h. Find the value of the coefficient of kinetic friction between the road and crate if the crate slides 50 m on the road in coming to rest. The initial speed of the crate is the same as the truck, 100 km/h.

\(\mu_k = 0.787\)

Use the kinematic equations to find the acceleration of the crate (with zero final speed).

\(0 = v_0^2 - 2 a \Delta s\)

\(\displaystyle a= \frac{v_0^2}{2 \Delta s}\)

Use \(F=ma\) find the net force, which equals the friction force.

\(f_k = \mu_k N = \mu_k mg = ma\)

So, solving for \(\mu_k\):

\(\displaystyle \mu_k = \frac{a}{g} = \frac{v_0^2}{2 g \Delta s} = 0.787\)

A car rounds an unbanked curve of radius \(65\) m. If the coefficient of static friction between the road and car is \(0.70\), what is the maximum speed at which the car can traverse the curve without slipping?

Slip is impeding when the centripetal force required for the turn (\(m v^2/r\)) equals the maximum static friction force (\(\mu_s N\)). On an unbanked curve, \(N=mg\).

So \(\displaystyle \frac{v^2}{r}= \mu_s g\), or

\(\displaystyle v = \sqrt{r mu_s g} = 21.1\) m/s

Suppose that the magnitude of the resistive force of the air on a \(50.0\) kg skydiver can be approximated by \(F_D=bv^2\) .

- Draw of free-body diagram of the skydiver when he is falling at terminal velocity.
- If the terminal velocity of the skydiver is found to be \(60.0\) m/s, what is the value of \(b\) (with its units)?

\(F_{net} = 0 =bv^2 - mg\)

\(b = \frac{mg}{v^2} = 0.136\) kg/m

*Last modified: Mon October 07 2024, 12:58 PM.*