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2 years ago

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Ezra is pulling a sled, filled with snow, by pulling on a rope attached to the sled. The rope makes an angle θ with respect to t

he horizontal ground, and the sled is being pulled at a constant speed. If the sled and snow have a total mass of m, the acceleration due to gravity is g, the magnitude of the normal force is FN, and the coefficient of kinetic friction between the sled and the ground is μk, what is the angle θ that the rope makes with the ground in terms of these five quantities? (Use the following as necessary: m, g, FN, and μk.)

1 answer:

svetoff [14.1K]2 years ago

8 0

Answer:

Explanation:

Given

rope makes an angle of $\theta$

Mass of sled and snow is m

Normal Force $=F_N$

applied Force is F

as Force is pulling in nature therefore normal reaction is given by

$F_N=mg-F\sin \theta$

Also $F\cos \theta =f_r$

$f_r=\mu _k\cdot F_N$

$f_r=\mu _k\cdot (mg-F\sin \theta )$

$F\cos \theta =\mu _kF_N$ -------1

$F\sin \theta =mg-F_N$ ---------2

Squaring 1 & 2 and then adding

$F^2=(\mu _kF_N)^2+(mg-F_N)^2$

$F=\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}$

Substitute value of F in 1

$cos\theta =\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}}$

$\theta =cos^{-1}(\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}})$

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1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

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Answer:

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Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

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Answer with Explanation:

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