# Tutor profile: Kevin T.

## Questions

### Subject: Physics (Newtonian Mechanics)

A block of mass $$ m=10 $$ $$ kg $$ rests on a perfectly horizontal surface with a coefficient of kinetic friction $$ \mu_{k}=0.2 $$. A force P acts on the block at an angle of 30 degrees with respect to the surface/horizon towards the upwards vertical direction. The block has a horizontal acceleration of $$ a_{x}=5 $$ $$ \frac{m}{s^2} $$ in the same direction as the horizontal component of the $$ P $$ force. Ignoring air resistance forces, what is the force P?

Newton’s second law of motion gives us the equation: $$ F = ma $$ Where $$ F $$ is the sum of the forces in a specific direction, $$ m $$ is the mass, and $$ a $$ is the acceleration in the same direction. $$ Vertical $$ $$ Forces $$ $$ (y-axis) $$ The vertical forces acting on the block are: the weight of the block $$ mg $$ downwards, the reactive normal force from the surface $$ N $$ upwards, and the vertical (upward) component of the force $$ P $$ which is $$ P\sin(30) $$. Since the block rests on the surface, the vertical acceleration $$ a_{y}=0 $$ and using Newton’s equation, we obtain: $$ P\sin(30)+N-mg=ma_{y} $$ $$ P\frac{1}{2}+N-10g=10(0) $$ $$ \frac{1}{2}P+N-10g=0 $$ We have the equation for the normal force $$ N $$ in terms of the force $$ P $$: $$ N=10g-\frac{1}{2}P $$ We’ll save that for later, now let’s look at the horizontal direction. $$ Horizontal $$ $$ Forces $$ $$ (x-axis) $$ The horizontal forces acting on the block are: the horizontal component of the force $$ P $$ which is $$ P\cos(30) $$ and the reactive friction force in the opposite direction, $$ F_{f}=\mu_{k}N $$. WIth the horizontal acceleration of $$ a_{x}=5 $$ $$ \frac{m}{s^2} $$, we use Newton’s equation again to obtain: $$ P\cos(30)-F_{f}=ma_{x} $$ $$ P\frac{\sqrt{3}}{2}-\mu_{k}N=10(5) $$ $$ \frac{\sqrt{3}}{2}P-(0.2)(10g-\frac{1}{2}P) = 50 $$ $$ \frac{\sqrt{3}}{2}P-2g+0.1P = 50 $$ $$ P=\frac{50+2g}{\frac{\sqrt{3}}{2}+0.1} $$ with $$ g=9.81 $$ $$ \frac{m}{s^2} $$, the force $$ P $$ is approximately: $$ P = 72 N $$

### Subject: Calculus

A curve on the xy-plane is defined by the function $$ f(x)=3x^2+4x-2 $$, determine: a) The slope of the curve at $$ x=2 $$ b) The nature of the curve's concavity. c) The area under the curve between $$ x=1 $$ and $$ x=4 $$.

a) The slope of the curve $$ f(x) $$ at $$ x=a $$ can be determined by taking its derivative $$ f'(x) $$ then finding its value at $$ x=a$$: $$ Slope=f'(a) $$ Thus we take the derivative of our function: $$ \frac{d}{dx}(3x^2+4x-2)=6x+4 $$ Our slope at $$ x=2 $$ is: $$ Slope=6(2)+4=12+4=16 $$ The slope of the curve at $$ x=2 $$ is 16. b) The nature of a curve $$ f(x) $$'s concavity at $$ x=a $$ can be determined by taking the second derivative and looking at its value at $$ x=a $$. So we have our derivative of $$ f(x) $$: $$ f'(x)=6x+4 $$ Taking the second derivative with respect to x again: $$ f''(x)=6 $$ A curve concave downwards when $$ f''(x) $$ is negative and concave upwards when $$ f''(x) $$ is positive. The double derivative of our function is a positive number (6) regardless of the value of x. So we can conclude that the curve always concave upwards independent of the value of x. c) The area under a curve f(x) between a and b is determined using the integral: $$ \int_a^b \mathrm{f(x)}\mathrm{d}x=[h(x)]^{b}_{a}=h(b)-h(a) $$ Where $$ h(x) $$ is the integral of $$ f(x) $$. So to determine the area under the curve $$ f(x)=3x^2+4x-2 $$ between $$ x=1 $$ and $$ x=4 $$, we solve the integral: $$ \int_1^4 \mathrm{(3x^2+4x-2)} \mathrm{d}x $$ Which gives us: $$ [x^3+2x^2-2x]^{4}_{1} $$ Solving gives: $$ (4)^3+2(4)^2-2(4)-[(1)^3+2(1)^2-2(1)] $$ $$ 64+2(16)-8-[1+2-2] $$ $$ 64+32-8-1 $$ $$ 87 $$ The area under the curve from $$ x=1 $$ to $$ x=4 $$ is 87.

### Subject: Algebra

Anne, Bridget, and Charlie are of very different ages. Charlie's current age is 12 times Anne's current age. Two years ago, Charlie's age was 5 times Bridget's age. In 4 years, Anne's age will be half of Bridget's age. How old are Anne, Bridget, and Calvin currently?

Since this is an algebraic questions, let's represent Anne, Bridget, and Charlie's age with variables: $$ a= $$ Anne's age $$ b= $$ Bridget's age $$ c= $$ Charlie's age Charlie's current age is 12 times Anne's current age. We can represent this with algebra to obtain: $$ First $$ $$ Equation $$ $$ c=12a $$ Two years ago, Charlie's age was 5 times Bridget's age. Using algebra again and then simplifying, we obtain: $$ Second $$ $$ Equation $$ $$ c-2=5(b-2) $$ $$ c-2=5b-10 $$ (Multiplying out the right-hand side of the equation) $$ c=5b-8 $$ (Adding 2 to both sides) Finally, in 4 years Anne's age will be half of Bridget's age. we can obtain our final equation once again with algebra and simplification: $$ Third $$ $$ Equation $$ $$ a+4=\frac{1}{2}(b+4) $$ $$ 2a+8=b+4 $$ (Multiplying both sides by 2) $$ 2a+4=b $$ (Subtracting both sides by 4) We now have three algebraic equations relating all three ages: $$ 1) $$ $$ c=12a $$ $$ 2) $$ $$ c=5b-8 $$ $$ 3) $$ $$ 2a+4=b $$ The first two equations are of Charlie's age in terms of Anne's age and Bridget's age, respectively. So we can equate the two and get a new equation in terms of the variables a and b: $$ 5b-8=12a $$ our third equations tells us that $$ b=2a+4 $$, therefore we can plug that in and solve for Anne's age $$ a $$: $$ 5(2a+4)-8=12a $$ $$ 10a+20-8=12a $$ (Multiplying out the parenthesis on the left-hand side) $$ 10a+12=12a $$ (Simplifying the left-hand side) $$ 12=2a $$ (Subtracting 10a from both sides) $$ a=6 $$ (Dividing both sides by 2) Anne is 6 years old. Knowing Anne's age, we plug that into the first equation to obtain Charlie's age: $$ c = 12(6) $$ $$ c = 72 $$ Charlie is 72 years old. Knowing Charlie's age, we plug that into the second equation to obtain Bridget's age: $$ 72=5b-8 $$ $$ 80=5b $$ (Adding 8 to both sides) $$ b=16$$ (Dividing both sides by 5) Bridget is 16 years old. Our final answer is: Anne is 4 years old, Bridget is 16 years old, and Charlie is 72 years old. We are done!

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