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.
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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
$$
\bar P = \frac{450}{15} = 30 \text{ W}
$$
1.11Determine the Direction of Static Friction on an Incline
An otherwise unforced block is stationary on a $30^\circ$ incline. Which direction does the static friction force on the block point?
Solution
The component of gravity parallel to the surface points down the incline. Without friction, that component would produce impending motion down the incline.
Static friction opposes that impending relative motion, so it acts up the incline.
For equilibrium, its required magnitude is $mg\sin 30^\circ$ as long as that value does not exceed the maximum available static friction. Static friction is not automatically equal to $\mu_sN$; $\mu_sN$ is only its maximum possible magnitude.
Difficulty
Integrated
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
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
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:
$$
\frac{1}{2}(0.50)v^2 = 1.0
$$
$$
0.25v^2 = 1.0
$$
$$
v^2 = 4
$$
$$
v = 2.0 \text{ m/s}
$$
4.5Test Whether Static Friction Is Sufficient
A $10$ kg block rests on a $30^\circ$ incline. The coefficient of static friction is $0.40$. Can static friction keep the block at rest? If not, explain why not. Use $g = 9.8\ \text{m/s}^2$.
Solution
Choose axes parallel and perpendicular to the incline. The force trying to make the block slide down the incline is
$$
f_{s,\mathrm{required}} = mg\sin 30^\circ
= (10)(9.8)(0.5) = 49\ \text{N}.
$$
The normal force is
$$
N = mg\cos 30^\circ \approx 84.9\ \text{N},
$$
so the largest possible static friction is
$$
f_{s,\max} = \mu_sN = (0.40)(84.9) \approx 34.0\ \text{N}.
$$
Because $49\ \text{N} > 34.0\ \text{N}$, static friction is not sufficient. The block cannot remain at rest and begins to slide down the incline. Notice that $f_s$ is not set equal to $\mu_sN$ unless the surfaces are at the threshold of slipping.