1.1Identify the SI Unit of Torque
What is the SI unit of torque?
Solution
Torque is force times distance, so its SI unit is
$$
\text{N} \cdot \text{m}
$$
This is equivalent to newton-meter.
Physics I Practice
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Difficulty
Direct Practice
1.2Classify Acceleration as a Scalar or Vector
Is acceleration a scalar or a vector?
Solution
Acceleration is a vector because it has both magnitude and direction.
So the correct classification is vector.
1.3Compute Average Velocity
A cyclist moves from $x = 15$ m to $x = 87$ m in $12$ s. What is the average velocity?
Solution
Use average velocity:
$$
\bar v = \frac{\Delta x}{\Delta t}
$$
The displacement is
$$
\Delta x = 87 - 15 = 72 \text{ m}
$$
So
$$
\bar v = \frac{72}{12} = 6 \text{ m/s}
$$
1.4Find Final Velocity with Constant Acceleration
An object starts with velocity $2$ m/s and accelerates at $3$ m/s$^2$ for $4$ s. What is its final velocity?
Solution
Use the constant-acceleration equation:
$$
v = v_0 + at
$$
Substitute the values:
$$
v = 2 + 3(4) = 14 \text{ m/s}
$$
1.5Find Position Under Constant Acceleration
An object starts at $x_0 = -3$ m with velocity $5$ m/s and acceleration $-2$ m/s$^2$. Where is it after $3$ s?
Solution
Use
$$
x = x_0 + v_0 t + \frac{1}{2}at^2
$$
Substitute the values:
$$
x = -3 + 5(3) + \frac{1}{2}(-2)(3^2)
$$
$$
x = -3 + 15 - 9 = 3 \text{ m}
$$
1.6Resolve a Vector into Components
A force of $20$ N acts at an angle of $60^\circ$ above the positive $x$-axis. Find its $x$- and $y$-components.
Solution
Use component formulas:
$$
F_x = F\cos\theta, \qquad F_y = F\sin\theta
$$
So
$$
F_x = 20\cos 60^\circ = 10 \text{ N}
$$
and
$$
F_y = 20\sin 60^\circ \approx 17.3 \text{ N}
$$
1.7State the Acceleration of Free Fall
If upward is positive, what is the acceleration of a dropped object near Earth?
Solution
Near Earth's surface, free-fall acceleration is approximately constant:
$$
a = -g \approx -9.8 \text{ m/s}^2
$$
The negative sign appears because the acceleration points downward.
1.8Use Newton's Second Law
A $6$ kg cart has a net horizontal force of $18$ N acting on it. What is its acceleration?
Solution
Apply Newton's second law:
$$
\sum F = ma
$$
So
$$
a = \frac{F}{m} = \frac{18}{6} = 3 \text{ m/s}^2
$$
1.9Find the Weight of an Object
What is the weight of a $12$ kg object near Earth's surface?
Solution
Weight is
$$
W = mg
$$
So
$$
W = 12(9.8) = 117.6 \text{ N}
$$
The force points downward.
1.10Compute Average Power
A machine does $450$ J of work in $15$ s. What is its average power?
Solution
Use average power:
$$
\bar P = \frac{W}{\Delta t}
$$
So
$$
\bar P = \frac{450}{15} = 30 \text{ W}
$$
Difficulty
Integrated Practice
2.1Find the Time of Flight for a Projectile
A ball is launched at $20$ m/s at an angle of $30^\circ$ above the horizontal and lands at the same height. How long is it in the air?
Solution
The vertical component of the launch velocity is
$$
v_{0y} = 20\sin 30^\circ = 10 \text{ m/s}
$$
For a projectile that lands at the same height,
$$
T = \frac{2v_{0y}}{g}
$$
So
$$
T = \frac{2(10)}{9.8} \approx 2.04 \text{ s}
$$
2.2Find Acceleration with Kinetic Friction
A $5$ kg block slides on a horizontal floor. It is pulled horizontally by an $18$ N force, and the coefficient of kinetic friction is $0.20$. What is the block's acceleration?
Solution
Because the pull is horizontal, the normal force is
$$
N = mg = 5(9.8) = 49 \text{ N}
$$
Kinetic friction is
$$
f_k = \mu_k N = 0.20(49) = 9.8 \text{ N}
$$
The net horizontal force is
$$
F_{net} = 18 - 9.8 = 8.2 \text{ N}
$$
So the acceleration is
$$
a = \frac{F_{net}}{m} = \frac{8.2}{5} = 1.64 \text{ m/s}^2
$$
2.3Use Energy to Find a Spring Speed
A $0.50$ kg block is attached to a spring with $k = 200$ N/m. The spring is compressed $0.10$ m and the surface is frictionless. What is the block's speed when the spring returns to equilibrium?
Solution
Use conservation of mechanical energy:
$$
\frac{1}{2}kx^2 = \frac{1}{2}mv^2
$$
Substitute the values:
$$
\frac{1}{2}(200)(0.10)^2 = \frac{1}{2}(0.50)v^2
$$
The left side is $1.0$ J, so
$$
1.0 = 0.25v^2
$$
$$
v^2 = 4
$$
$$
v = 2.0 \text{ m/s}
$$
2.4Find the Final Speed in a Perfectly Inelastic Collision
A $2$ kg cart moving $3$ m/s east sticks to a $4$ kg cart at rest. What is the final velocity of the combined carts?
Solution
Use conservation of momentum:
$$
m_1v_{1i} + m_2v_{2i} = (m_1+m_2)v_f
$$
Substitute:
$$
(2)(3) + (4)(0) = 6v_f
$$
$$
6 = 6v_f
$$
$$
v_f = 1 \text{ m/s}
$$
The direction is east.
2.5Find Angular Acceleration from Torque
A force of $12$ N is applied perpendicular to a wrench that is $0.40$ m long. If the moment of inertia about the rotation axis is $0.80$ kg m$^2$, what is the angular acceleration?
Solution
First find the torque:
$$
\tau = rF = 0.40(12) = 4.8 \text{ N m}
$$
Then use rotational dynamics:
$$
\sum \tau = I\alpha
$$
So
$$
\alpha = \frac{\tau}{I} = \frac{4.8}{0.80} = 6 \text{ rad/s}^2
$$
2.6Find the Centripetal Force
A $0.50$ kg stone tied to a string of radius $2.0$ m moves in a circle at $6.0$ m/s. What inward force is required?
Solution
Use the centripetal-force formula:
$$
F_c = \frac{mv^2}{r}
$$
Substitute:
$$
F_c = \frac{(0.50)(6.0^2)}{2.0}
$$
$$
F_c = \frac{(0.50)(36)}{2.0} = 9 \text{ N}
$$
The force must point inward, toward the center of the circle.
2.7Find the Period of a Mass-Spring System
A $0.80$ kg mass is attached to a spring with $k = 50$ N/m. What is the period of the oscillation?
Solution
For a mass-spring system,
$$
T = 2\pi\sqrt{\frac{m}{k}}
$$
Substitute the values:
$$
T = 2\pi\sqrt{\frac{0.80}{50}}
$$
$$
T \approx 2\pi(0.126) \approx 0.79 \text{ s}
$$
2.8Find the Center of Mass on a Line
Two masses lie on the $x$-axis: a $2$ kg mass at $x = 0$ m and a $6$ kg mass at $x = 4$ m. Where is the center of mass?
Solution
Use the one-dimensional center-of-mass formula:
$$
x_{cm} = \frac{\sum m_i x_i}{\sum m_i}
$$
So
$$
x_{cm} = \frac{(2)(0) + (6)(4)}{2+6}
$$
$$
x_{cm} = \frac{24}{8} = 3 \text{ m}
$$
Difficulty
Applied Problems
3.1Find How Far a Horizontal Projectile Travels
A ball is launched horizontally at $14$ m/s from a cliff that is $20$ m high. How far from the base of the cliff does it land?
Solution
First find the time to fall $20$ m vertically:
$$
20 = \frac{1}{2}gt^2
$$
So
$$
t = \sqrt{\frac{40}{9.8}} \approx 2.02 \text{ s}
$$
Horizontal speed stays constant, so
$$
x = v_xt = 14(2.02) \approx 28.3 \text{ m}
$$
3.2Find Acceleration for a Pulled Block on a Rough Surface
A $10$ kg crate is pulled by a $40$ N force at an angle of $25^\circ$ above the horizontal. The coefficient of kinetic friction is $0.30$. What is the crate's acceleration?
Solution
Resolve the pulling force into components:
$$
F_x = 40\cos 25^\circ \approx 36.3 \text{ N}
$$
$$
F_y = 40\sin 25^\circ \approx 16.9 \text{ N}
$$
The upward component reduces the normal force:
$$
N = mg - F_y = 10(9.8) - 16.9 = 81.1 \text{ N}
$$
So the kinetic friction is
$$
f_k = \mu_k N = 0.30(81.1) \approx 24.3 \text{ N}
$$
The net horizontal force is
$$
F_{net} = 36.3 - 24.3 = 12.0 \text{ N}
$$
Therefore,
$$
a = \frac{12.0}{10} = 1.20 \text{ m/s}^2
$$
3.3Find the Final Velocity After a Collision
A $1.5$ kg cart moves east at $4$ m/s and hits a $0.5$ kg cart moving west at $1$ m/s. The carts stick together. What is their final velocity?
Solution
Choose east as positive. Then the initial momentum is
$$
p_i = (1.5)(4) + (0.5)(-1) = 6 - 0.5 = 5.5 \text{ kg m/s}
$$
The total mass after the collision is
$$
1.5 + 0.5 = 2.0 \text{ kg}
$$
So
$$
v_f = \frac{p_i}{m_{total}} = \frac{5.5}{2.0} = 2.75 \text{ m/s}
$$
The direction is east.
3.4Find the Support Force on a Beam
A uniform $4.0$ m beam weighs $200$ N. It is supported at both ends, and a $100$ N sign hangs $1.0$ m from the left end. What is the force on the left support?
Solution
Let the left support force be $F_L$ and the right support force be $F_R$.
First use torque equilibrium about the left end:
$$
F_R(4.0) - 200(2.0) - 100(1.0) = 0
$$
So
$$
4F_R = 500
$$
$$
F_R = 125 \text{ N}
$$
Now use vertical force balance:
$$
F_L + F_R = 200 + 100 = 300
$$
Thus
$$
F_L = 300 - 125 = 175 \text{ N}
$$
3.5Use Rolling Without Slipping
A wheel of radius $0.30$ m rolls without slipping at $4.5$ m/s for $6.0$ s. How many radians does it rotate?
Solution
For rolling without slipping,
$$
v = r\omega
$$
So
$$
\omega = \frac{v}{r} = \frac{4.5}{0.30} = 15 \text{ rad/s}
$$
Angular displacement is
$$
\theta = \omega t = 15(6.0) = 90 \text{ rad}
$$
Difficulty
Challenge / Synthesis
4.1Relate Spring Energy to Gravitational Height
A $2.0$ kg block starts at rest on a frictionless track. A spring with $k = 300$ N/m launches it so that it rises $1.5$ m before stopping. How much was the spring compressed?
Solution
Use conservation of mechanical energy. The spring's initial energy becomes gravitational potential energy at the top:
$$
\frac{1}{2}kx^2 = mgh
$$
Substitute the values:
$$
\frac{1}{2}(300)x^2 = (2.0)(9.8)(1.5)
$$
$$
150x^2 = 29.4
$$
$$
x^2 = 0.196
$$
$$
x \approx 0.44 \text{ m}
$$
4.2Find the Friction Force on a Leaning Ladder
A uniform $5.0$ m ladder weighs $200$ N and leans against a frictionless wall at an angle of $60^\circ$ above the floor. What friction force must the floor provide to keep the ladder at rest?
Solution
Because the wall is frictionless, the wall exerts only a horizontal force on the top of the ladder. Let that force be $F_W$.
Take torques about the bottom of the ladder so the floor forces do not appear.
The wall force acts at the top with lever arm
$$
5.0\sin 60^\circ
$$
The weight acts at the center of the ladder, with horizontal lever arm
$$
2.5\cos 60^\circ
$$
Torque balance gives
$$
F_W(5.0\sin 60^\circ) = 200(2.5\cos 60^\circ)
$$
So
$$
F_W = \frac{200(2.5)(0.5)}{5.0(0.866)} \approx 57.7 \text{ N}
$$
Horizontal force balance means the floor friction must match this force in magnitude:
$$
f_s \approx 57.7 \text{ N}
$$
4.3Show That Orbital Speed Does Not Depend on Satellite Mass
A satellite moves in a circular orbit of radius $r$ around a planet of mass $M$. Use the formulas in the note to show that the satellite's mass cancels, and give the orbital speed.
Solution
For a circular orbit, gravity supplies the centripetal force:
$$
\frac{GMm}{r^2} = m\frac{v^2}{r}
$$
The satellite mass $m$ appears on both sides, so it cancels:
$$
\frac{GM}{r^2} = \frac{v^2}{r}
$$
Multiply by $r$:
$$
\frac{GM}{r} = v^2
$$
Therefore,
$$
v = \sqrt{\frac{GM}{r}}
$$
The orbital speed does not depend on the satellite mass.
4.4Find the Speed at Equilibrium in Simple Harmonic Motion
A $0.50$ kg mass on a spring with $k = 200$ N/m is released from rest at an amplitude of $0.10$ m. What is its speed when it passes through equilibrium?
Solution
The total energy in ideal SHM is
$$
E = \frac{1}{2}kA^2
$$
So
$$
E = \frac{1}{2}(200)(0.10)^2 = 1.0 \text{ J}
$$
At equilibrium, all of that energy is kinetic:
$$
\frac{1}{2}mv^2 = 1.0
$$
Substitute $m = 0.50$ kg:
$$
\frac{1}{2}(0.50)v^2 = 1.0
$$
$$
0.25v^2 = 1.0
$$
$$
v^2 = 4
$$
$$
v = 2.0 \text{ m/s}
$$