\text{A $2.0 \mathrm{~kg}$ block is thrown upward from the ground. At what height above the ground will }

\text{the gravitational potential energy of the Earth-block system have increased by $490 \mathrm{~J}$ ?}

\text{A) $12 \mathrm{~m}$}\\ \text{B) $50 \mathrm{~m}$}\\ \text{C) $25 \mathrm{~m}$}\\ \text{D) $8.0 \mathrm{~m}$}\\ \text{E) $18 \mathrm{~m}$}

\text{A $2.0 \mathrm{~kg}$ block is thrown upward from the ground. At what height above the ground will }

\text{the gravitational potential energy of the Earth-block system have increased by $490 \mathrm{~J}$ ?}

\text{A) $12 \mathrm{~m}$}\\ \text{B) $50 \mathrm{~m}$}\\ \text{C) $25 \mathrm{~m}$}\\ \text{D) $8.0 \mathrm{~m}$}\\ \text{E) $18 \mathrm{~m}$}

\Delta U=mgy

\text{where }y \text{ is measure from the ground as reference.}

\displaystyle y=\frac{\Delta U}{mg}=\frac{490}{2.0\times 9.8}\approx 25\text{ m}

\text{Answer C)}

\text{An ideal spring with a $20 \mathrm{~N} / \mathrm{m}$ spring constant is compressed by a $10 \mathrm{~N}$ force. The potential }

\text{A) $0.50 \mathrm{~J}$}\\ \text{B) $2.5 \mathrm{~J}$}\\ \text{C) $5.0 \mathrm{~J}$}\\ \text{D) $10 \mathrm{~J}$}\\ \text{E) $200 \mathrm{~J}$}

\text{energy stored in the spring is: }

\text{An ideal spring with a $20 \mathrm{~N} / \mathrm{m}$ spring constant is compressed by a $10 \mathrm{~N}$ force. The potential }

\text{A) $0.50 \mathrm{~J}$}\\ \text{B) $2.5 \mathrm{~J}$}\\ \text{C) $5.0 \mathrm{~J}$}\\ \text{D) $10 \mathrm{~J}$}\\ \text{E) $200 \mathrm{~J}$}

\text{Answer B)}

\text{energy stored in the spring is: }

\displaystyle U=\frac{1}{2}k x_\text{}^2

\text{The elastic potential energy is:}

\text{(When the spring is relaxed, its energy is 0)}

\displaystyle F=k x

\text{(magnitude)}

\displaystyle U=\frac{1}{2}\frac{F^2}{k} =\frac{10^2}{2\times20}=2.5 \text{ J}

\text{A $2.0 \mathrm{~kg}$ object is connected to one end of an unstretched spring which is attached to the ceiling }

\text{A) $0.80 \mathrm{~m}$}\\ \text{B) $0.10 \mathrm{~m}$}\\ \text{C) $0.40 \mathrm{~m}$}\\ \text{D) $0.20 \mathrm{~m}$}\\ \text{E) $0.50 \mathrm{~m}$}

\text{by the other end and then the object is allowed to drop. The spring constant of the spring }

\text{is $196 \mathrm{~N} / \mathrm{m}$. How far does it drop before coming to rest momentarily?}

m

m

\text{reference}

k

\text{(not compressed)}

\text{A $2.0 \mathrm{~kg}$ object is connected to one end of an unstretched spring which is attached to the ceiling }

\text{by the other end and then the object is allowed to drop. The spring constant of the spring }

\text{is $196 \mathrm{~N} / \mathrm{m}$. How far does it drop before coming to rest momentarily?}

\text{We take the same reference for the potential energy and the elastic one.}

\text{To avoid problems, take it at the position }\\ \text{where the spring is not stretched.}

\text{The mechanical energy is conserved:}

K_\text{i}+U_\text{i}=K_\text{f}+U_\text{f}

\displaystyle \frac{1}{2}mv_\text{i}^2+mgy_0+\frac{1}{2} k x_\text{i}^2=\frac{1}{2}mv_\text{f}^2+mgy+\frac{1}{2} k x_\text{f}^2

x_\text{i}=y_0=0

v_\text{i}=v_\text{f}=0

x_\text{f}=y

(\text{reference})

(\text{at rest})

\text{(same distance)}

\displaystyle 0=mgy+\frac{1}{2} k y^2

\displaystyle 0=mgy+\frac{1}{2} k y^2

\displaystyle y=-\frac{2mg}{k}=-\frac{2\times2\times9.8}{196}

\Rightarrow

=-0.20\text{ m}

\text{Answer D)}

\text{(not compressed)}

\text{An ideal spring (compressed by $7.00 \mathrm{~cm}$ and initially at rest,) fires a 15.0 $\mathrm{g}$ block horizontally }

\text{across a frictionless table top. The spring has a spring constant of $20.0 \mathrm{~N} / \mathrm{m}$. }

\text{The speed of the block as it leaves the spring is: }

\text{A) $2.56 \mathrm{~m} / \mathrm{s}$}\\ \text{B) $1.90 \mathrm{~m} / \mathrm{s}$}\\ \text{C) $3.64 \mathrm{~m} / \mathrm{s}$}\\ \text{D) $8.12 \mathrm{~m} / \mathrm{s}$}\\ \text{E) $5.25 \mathrm{~m} / \mathrm{s}$}

\text{An ideal spring (compressed by $7.00 \mathrm{~cm}$ and initially at rest,) fires a 15.0 $\mathrm{g}$ block horizontally }

\text{across a frictionless table top. The spring has a spring constant of $20.0 \mathrm{~N} / \mathrm{m}$. }

\text{The speed of the block as it leaves the spring is: }

m

m

7.00 \text{ cm}

\text{The external forces }\vec{T}\text{ and }\vec{F}_N \text{ are not working}

\text{The mechanical energy of (spring+mass) is \textcolor{red}{conserved}}

K_\text{i}+U_\text{i}=K_\text{f}+U_\text{f}

\displaystyle \frac{1}{2}mv_\text{i}^2+\frac{1}{2} k x_\text{i}^2=\frac{1}{2}mv_\text{f}^2+\frac{1}{2} k x_\text{f}^2

v_\text{i}=0

x_\text{f}=0

(\text{at rest})

\text{(not stretched)}

\text{We have }

\text{ and }

\displaystyle \frac{1}{2} k x_\text{i}^2=\frac{1}{2}mv_\text{f}^2

\displaystyle v=x_\text{i}\sqrt{\frac{k}{m}}

\displaystyle v=0.07\times\sqrt{\frac{20}{15 \times 10^{-3}}}\approx2.56\text{ m/s}

\text{Answer A)}

\text{(kinetic and elastic energies) }

\text{An object of mass $m$, attached to a light cord of length $L$, is held horizontally from a fixed}

\text{ support as shown in Fig 1 . The object is then released from rest.}

\text{ What is the tension force in the cord when the object is at the lowest point of its swing?}

\text{A) $2 \mathrm{ mg}$}\\ \hspace{-2mm}\text{B) $\mathrm{ mg}$}\\ \text{C) $3 \mathrm{ mg}$}\\ \hspace{2mm}\text{D) $\mathrm{ mg} / 2$}\\ \text{E) $\mathrm{ mgL}$}

\text{An object of mass $m$, attached to a light cord of length $L$, is held horizontally from a fixed}

\text{ support as shown in Fig 1 . The object is then released from rest.}

\text{ What is the tension force in the cord when the object is at the lowest point of its swing?}

\Delta E_{mec}=W(\vec{T})

\text{since }\vec{T} \text{ is continously perpendicular to the path,}\\ \text{ its work is therefore }0

\text{The mechanical energy is therefore \textcolor{red}{conserved}}

K_\text{i}+U_\text{i}=K_\text{f}+U_\text{f}

\displaystyle \frac{1}{2}mv_\text{i}^2+mgy=\frac{1}{2}mv_\text{f}^2

\text{The reference for the potential energy is taken at the lowes point}

y=L

v_\text{i}=0

\text{ and }

\Rightarrow

\displaystyle v_\text{f}=\sqrt{2gL}

\text{(system is earth+mass)}

\text{at the lowest point, we have}

m\vec{g}+\vec{T}=m\vec{a}

\text{where the acceleration is centripetal. By projecting on the radial axis, we get:}

\displaystyle -m{g}+{T}=m{a}=m\frac{v^2}{R}

\Rightarrow

\displaystyle {T}=mg+m\frac{v^2}{R}=mg+2mgL/L=3mg

\text{Answer C)}

\displaystyle {T}=3mg

\text{A block of mass $2.0 \mathrm{~kg}$ is initially moving to the right on a horizontal frictionless surface at }

\text{ a speed $5.0 \mathrm{~m} / \mathrm{s}$. It then compresses a spring of spring constant $100 \mathrm{~N} / \mathrm{m}$.}

\text{ At the instant when the kinetic energy of the block is equal to the potential energy of the spring,}

\text{ the spring is compressed a distance of:}

\text{A) $0.25 \mathrm{~m}$}\\ \text{B) $0.50 \mathrm{~m}$}\\ \hspace{-2mm}\text{C) $1.0 \mathrm{~m}$}\\ \text{D) $0.75 \mathrm{~m}$}\\ \text{E) $0.10 \mathrm{~m}$}

\text{A block of mass $2.0 \mathrm{~kg}$ is initially moving to the right on a horizontal frictionless surface at }

\text{ a speed $5.0 \mathrm{~m} / \mathrm{s}$. It then compresses a spring of spring constant $100 \mathrm{~N} / \mathrm{m}$.}

\text{ At the instant when the kinetic energy of the block is equal to the potential energy of the spring,}

\text{ the spring is compressed a distance of:}

{m}

{m}

?

k

\text{initial}

\text{final}

\text{The mechanical energy of the system mass+spring is \textcolor{red}{conserved.}}

K_\text{i}+U_\text{i}=K_\text{f}+U_\text{f}

\text{the initial elastic potential energy is }U_\text{i}=0

K_\text{f}=U_\text{f}\Rightarrow K_\text{f}+U_\text{f}=2U_\text{f}

\dots

K_\text{i}=2U_\text{f}

\Rightarrow

\displaystyle \frac{1}{2}mv_\text{i}^2= k x_\text{f}^2

\displaystyle x_\text{f}=\sqrt{\frac{m}{2k}}v_\text{i}

\displaystyle x_\text{f}=\sqrt{\frac{2.0}{2\times100}}\times 5.0=0.50 \text{ m}

\displaystyle x_\text{f}=0.50 \text{ m}

\text{Answer B)}

\Rightarrow

\Rightarrow

\Rightarrow

\Rightarrow

\text{A $3.00 \mathrm{~kg}$ block is dropped from a height of $40 \mathrm{~cm}$ onto a spring of spring constant $\mathrm{k}$ (see Fig 2).}

\text{ If the maximum distance the spring is compressed $=0.130 \mathrm{~m}$, find $\mathrm{k}$.}

\text{A) $1840 \mathrm{~N} / \mathrm{m}$}\\ \hspace{-2mm}\text{B) $980 \mathrm{~N} / \mathrm{m}$}\\ \hspace{-2mm}\text{C) $490 \mathrm{~N} / \mathrm{m}$}\\ \text{D) $1250 \mathrm{~N} / \mathrm{m}$}\\ \text{E) $2800 \mathrm{~N} / \mathrm{m}$}

\text{A $3.00 \mathrm{~kg}$ block is dropped from a height of $40 \mathrm{~cm}$ onto a spring of spring constant $\mathrm{k}$ (see Fig 2).}

\text{ If the maximum distance the spring is compressed $=0.130 \mathrm{~m}$, find $\mathrm{k}$.}

\text{system=earth+mass+spring}

\text{mechanical energy is \textcolor{red}{conserved}}

K_\text{i}+U_\text{i}^p+U_\text{i}^e=K_\text{f}+U_\text{f}^p+U_\text{f}^e

\displaystyle \frac{1}{2}mv_\text{i}^2+mgy_0+\frac{1}{2} k x_\text{i}^2=\frac{1}{2}mv_\text{f}^2+mgy+\frac{1}{2} k x_\text{f}^2

\displaystyle mg(y_0-y)=\frac{1}{2} k x_\text{f}^2

(\text{the final speed is }0)

y_0-y \text{ is the total distance: }y_0-y=d+x_\text{f}

\displaystyle k=\frac{2mg(d+x_\text{f})}{x_\text{f}^2}

\displaystyle k=\frac{2\times 3.00\times 9.8\times(0.40+0.130)}{0.130^2}\approx1844 \text{ N/m}

\text{Ansewr A)}

m

m

x

\text{Figure 2}

d

\text{A projectile of mass $0.20 \mathrm{~kg}$ is fired with an initial speed of $20 \mathrm{~m} / \mathrm{s}$ at an angle of 60 degrees}

\text{ above the horizontal. The kinetic energy of the projectile at its highest point is:}

\hspace{-1mm}\text{A) $0 \mathrm{~J}$}\\ \text{B) $40 \mathrm{~J}$}\\ \text{C) $30 \mathrm{~J}$}\\ \hspace{1mm}\text{D) $5.0 \mathrm{~J}$}\\ \text{E) $10 \mathrm{~J}$}

\text{A projectile of mass $0.20 \mathrm{~kg}$ is fired with an initial speed of $20 \mathrm{~m} / \mathrm{s}$ at an angle of 60 degrees}

\text{ above the horizontal. The kinetic energy of the projectile at its highest point is:}

\text{Conservation of the mechanical energy:}

K_\text{i}+U_\text{i}=K_\text{f}+U_\text{f}

\displaystyle \frac{1}{2}mv_\text{i}^2+mgy_0=K_\text{f}+mgy

\text{The maximum height }\displaystyle H=\frac{v_i^2 \sin(\theta)^2}{2g}

(\text{Formula sheet})

\displaystyle \frac{1}{2}mv_\text{i}^2-mgH=K_\text{f}

\displaystyle \frac{1}{2}mv_\text{i}^2 (1-\sin^2\theta)=K_\text{f}

\displaystyle K_\text{f}=\frac{1}{2}mv_\text{i}^2 \cos^2\theta

\displaystyle K_\text{f}=\frac{1}{2} 0.20\times20^2\times \frac{1}{2^2}=10 \text{ J}

\text{Answer E)}

\Rightarrow

\Rightarrow

\Rightarrow

\Rightarrow

\text{A $2.2 \mathrm{~kg}$ block starts from rest on a rough inclined plane that makes an angle of $25^{\circ}$ with the}

\text{horizontal. The coefficient of kinetic friction is 0.25 . As the block slides $2.0 \mathrm{~m}$ down the plane, }

\text{the mechanical energy of the Earth-block system changes by:}

\text{A) $-11 \mathrm{~J}$}\\ \hspace{-2mm}\text{B) $+0 \mathrm{~J}$}\\ \hspace{0mm}\text{C) $+9.8 \mathrm{~J}$}\\ \text{D) $-18 \mathrm{~J}$}\\ \hspace{1mm}\text{E) $-9.8 \mathrm{~J}$}

\text{A $2.2 \mathrm{~kg}$ block starts from rest on a rough inclined plane that makes an angle of $25^{\circ}$ with the}

\text{horizontal. The coefficient of kinetic friction is 0.25 . As the block slides $2.0 \mathrm{~m}$ down the plane, }

\text{the mechanical energy of the Earth-block system changes by:}

\Delta E_{\mathrm{mec}}+\Delta E_{\mathrm{th}}=0

\Delta E_{\mathrm{int}}=0

W(\vec{F}_N)=0

\text{The external force is }\vec{F}_N, \text{ so}

\Delta E_{\mathrm{th}}=f d=\mu_k N d=\mu_kmgd \cos\theta

\vec{F}_N

m\vec{g}

\vec{f}

\Delta E_{\mathrm{mec}}=-\Delta E_{\mathrm{th}}=-\mu_kmgd \cos\theta

=-0.25\times 2.2\times9.8\times2\times\cos(25^0)

=-0.25\times 2.2\times9.8\times2\times\cos(25^0)

=-9.8 \text{ J}

\text{Answer E)}

\text{A $6.0 \mathrm{~kg}$ box starts up a 30 degrees incline with $158 \mathrm{~J}$ of kinetic energy. How far will it slide up }

\text{the incline if the coefficient of kinetic friction between box and incline is 0.40 ?}

\text{A) $5.2 \mathrm{~m}$}\\ \text{B) $2.2 \mathrm{~m}$}\\ \text{C) $1.2 \mathrm{~m}$}\\ \text{D) $4.2 \mathrm{~m}$}\\ \text{E) $3.2 \mathrm{~m}$}

\text{A $6.0 \mathrm{~kg}$ box starts up a 30 degrees incline with $158 \mathrm{~J}$ of kinetic energy. How far will it slide up }

\text{the incline if the coefficient of kinetic friction between box and incline is 0.40 ?}

\Delta E_{\mathrm{mec}}+\Delta E_{\mathrm{th}}=0

\Delta K+\Delta U +\Delta E_{\mathrm{th}}=0

\Delta E_{\mathrm{th}}=f d=\mu_k N d=\mu_kmgd \cos\theta

\Delta K=K_\text{f}-K_\text{i}=0-K_\text{i}=-158 \text{ J}

\Delta U =mg(y-y_0)=mgd \sin\theta

\Delta K+\Delta U +\Delta E_{\mathrm{th}}=0

\Rightarrow

\Delta K+mg h+\mu_kmgd \cos\theta=0

\displaystyle d=-\frac{\Delta K}{mg(\sin\theta+\mu_k \cos\theta)}=\frac{158}{6.0\times 9.8\times(\sin30^0+0.40\cos30^0)}\approx3.2 \text{ m}

\Rightarrow

\Delta K+mg d \sin\theta+\mu_kmgd \cos\theta=0

\text{Conservation of energy}

\text{Answer E)}

\begin{aligned} I=I_{\text {hoop }}+ & I_{\text {rod }} \\ & I_{\text {hoop }}=\frac{1}{2} \mathrm{mR}^2 \end{aligned}