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Kevin T.
B.S in Engineering Mechanics, College tutoring experience (1 year), speaks 3 languages
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Physics (Newtonian Mechanics)
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Question:

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?

Kevin T.

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: $$P = 72 N$$

Calculus
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Question:

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 $$x=4$$.

Kevin T.

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.

Algebra
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Question:

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?

Kevin T.

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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