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# Tutor profile: Matt M.

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Matt M.
Physics Ph.D. and TA for six years
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A 2 kilogram block slides from rest down a 2 meter tall ramp, at an angle of 25 degrees above the ground. The coefficient of kinetic friction $$\mu_k$$ between the block and the ramp is 0.3. What is the block's speed at the bottom of the ramp?

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Matt M.

Here we consider the energy of the earth-block system. The initial gravitational potential energy is $(U_i=mgh_i=(2 kg)(9.8 m/s^2)(2 m)=39.2 J$) and the final gravitational potential energy is $(U_f=mgh_f=(2 kg)(9.8 m/s^2)(0 m)=0 J.$)Since the block starts at rest, the initial kinetic energy is 0 J, and the final kinetic energy will be $(K_f=\frac{1}{2}mv_f^2$)where $$v_f$$ is the final speed we want to calculate. Additionally, energy is not conserved in this system, but some energy is removed by the external work done by the friction of the ramp. This work is given by $$W_f=\overrightarrow{F_f}\cdot \overrightarrow{d}$$, where $$\overrightarrow{d}$$ is the block's displacement and$$\overrightarrow{F_f}$$ is the force of friction. The force is in the opposite direction of the displacement, so the work is negative (this makes sense, as we have already said energy is being removed from the system), and the two vectors are parallel, so the dot product is simply the product of their magnitudes. The magnitude of the force of friction is $(F_f = \mu_k F_N= \mu_k m g cos(\theta) = (0.3)(2 kg)(9.8 m/s^2)(cos(25 ^{\circ}) = 5.33 N$)where $$F_N$$ is the normal force, and the magnitude of the displacement is $( d=\frac{h}{sin(\theta)} = \frac{2m}{sin(25 ^{\circ})} = 4.73 m$) making the work done by friction $( W_f = -F_f \cdot d = -(5.33 N)(4.73 m) = -25.21 J.$) Now we put this all together: $( U_i + K_i +W_f = U_f + K_f$)$( 39.2 J + 0 J - 25.21 J = 0 J + \frac{1}{2}mv_f^2$)$( \frac{1}{2}(2 kg)v_f^2 = 13.99 J$)$(v_f=3.74 m/s$)

### Subject:Physics (Electricity and Magnetism)

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Question:

Using Gauss's Law, find the electric field produced by a sphere of radius $$R$$, with a charge $$Q$$ uniformly distributed through the volume, as a function of the distance $$r$$ from the center of the sphere.

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Matt M.

Gauss's Law tells us that the electric flux through an arbitrary closed surface (a so-called "Gaussian" surface) is proportional to the charge contained within the surface: $(\Phi_{e}=\frac{Q_{enc}}{\varepsilon_0}$) where $$\varepsilon_0 \approx 8.85 \times 10^{-12} F \cdot m^{-1}$$ is a fundamental constant called the permittivity of free space. Since we have a spherical system, we choose a spherical Gaussian surface centered at the center of the charged sphere - this will simplify things greatly. The definition of electric flux through a surface $$S$$ is $( \Phi_e = \int_{S} \overrightarrow{E} \cdot d\overrightarrow{A}$), but for a spherical surface and a spherical system, the electric field $$\overrightarrow{E}$$ and the area element $$d\overrightarrow{A}$$ are parallel, pointing directly away from the sphere's center, and the field is constant at a given radius, so we do not need to worry about the integral and instead simply multiply by the Gaussian surface's area: $( \Phi_e = E*A_{S} = E*(4\pi r^2) = \frac{Q_{enc}}{\varepsilon_0}$) So $( E=\frac{Q_{enc}}{4\pi r^2\varepsilon_0}$) and now all we need to find is $$Q_{enc}$$, and there are two possible cases: Case 1: $$r>R$$ Here $$Q_{enc}=Q$$ (since all the charge is inside the Gaussian surface) and $( E=\frac{Q}{4\pi r^2\varepsilon_0}$) Note that this simply looks the same as if all the charge was concentrated as a point charge at $$r=0$$. Case 2: $$r \leq R$$ Here not all of the charge $$Q$$ is inside the Gaussian surface. Instead, because the charge is uniformly distributed, we can use a ratio of the volume of the Gaussian sphere to the physical sphere: $( \frac{Q_{enc}}{Q} = \frac{V_{enc}}{V}=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{r^3}{R^3} \Longrightarrow Q_{enc}=Q\frac{r^3}{R^3}$) Substitute this into the equation for $$E$$ derived above: $( E=\frac{Q}{4\pi r^2\varepsilon_0} \frac{r^3}{R^3} = \frac{Q}{4\pi R^3\varepsilon_0}r$) Now we have the expression for the electric field both inside and outside of the sphere.

### Subject:Calculus

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Question:

Find the indefinite integral $$\int x^2 sin(x) dx$$ using integration by parts.

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Matt M.

In integration by parts, we recognize an integrand that is the product of an easy-to-differentiate function $$u$$ and an easy-to-integrate function $$v'$$, and use the relationship $$\int uv' = uv-\int u'v$$, which is derived from the chain rule. Here we observe that $$u=x^2$$ is easy to differentiate, and $$v'=sin(x)$$ is easy to integrate, and we get $$u'=2x$$ and $$v=-cos(x)$$, respectively. (For now we neglect the constant of integration, which only needs to be added at the end.) Then by substituting into the above relationship, we obtain $( \int x^2 sin(x) dx = -x^2cos(x) -\int 2x (-cos(x))dx = -x^2cos(x) + 2\int xcos(x)dx$). Here the second integral $$\int xcos(x)dx$$ can again be solved using integration by parts, using $$u=x$$ and $$v'=cos(x)$$, corresponding to $$u'=1$$ and $$v=sin(x)$$. This gives $(\int xcos(x)dx = xsin(x) -\int sin(x)dx = xsin(x) +cos(x) \space \space (+C)$), where the second integration is simple. Substituting this back into our above solution gives the final answer: $( \int x^2 sin(x) dx = -x^2cos(x) +2xsin(x) +2cos(x)+C$)

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