Physics 20 Unit B Assessment

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Jan 9, 2024

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Physics 20: Unit B Dynamics - also the study of motion. /28 At the beginning of this unit, you were given some “I can” statements and other “Questions” as part of “Questions You Should Be Able to Answer For This Unit” document. Your task for this assessment is to find an example question (from any source) that requires the information in each “Question” listed in that document. You’ll then explain how to answer each question, showing all your work, and ultimately, answer the question. You’ll be graded on each question with the following criteria: 1 point Explain how the question relates to the “Questions You Should Be Able to Answer For This Unit” question. 1 point Correct final answer for the question 2 points Relevant formula and work shown, or relevant concepts explained for how to get the answer Some questions might answer multiple “Questions” in this document. While you can use the same question more than once, you will need to re-explain how the question answers that addresses “Question” Remember, this assignment is worth 20% of your grade for this unit, so put a considerable amount of effort into this assignment.

1. Using Newton’s first law of motion, how can you describe an object’s state of rest? -A block of 80kg is supported by two cables with tension T1 and T2. Calculate T1 and T2. -When we solve this question. we assume all of the forces are balanced because the object is at rest which is what the first law of motion is (I have solved the question below under the "vector components" topic).

2. Using Newton’s second law of motion, what is the relationship between force, mass and acceleration? - A team of three kids is competing with other teams in a game of sliding pebbles on a frictionless table. The team whose all three rocks fall off the table at the same time wins. One of the three kids is holding a pebble of mass m, while the second pebble has twice the mass, and the third pebble is one third the first one. If the kid with the pebble of mass m slides it with a force of 5 N, what force should the second and third kids apply in order for all the pebbles to arrive at the edge of the table at the same time? -This question covers the relationship between the motion of objects, and their mass and the forces they experience. In order to solve it, we need to consider the proportionality of force, mass and acceleration, which Newton laid out in his second law of motion. -Given: F1=5N, m1=m, m2= 2m, m3, 1/3m, a1=a2=a3=a -Required: F2=? F3=? -Solution: F1=5N=m1xa1, 5N=ma, F2=m2xa2= 2ma = 2xF1= 2x5N=10N F3=m3xa3= 1/3ma = 1/3x5N = 5/3N -So, the second one has to apply 10N, and the third one has to apply 5/3N

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3. Using Newton’s third law of motion, how can you describe the interaction between two objects, remembering that the two forces (equal in magnitude, but opposite in direction) are not actually acting on the same object? -Three boxes, A(8.00kg), B(6.00kg), and C(12.0kg) are positioned next to each other, from left to right, on a lab table. A student pushes box A to the right causing all the boxes to accelerate at 2.82 m/s2 right. Assuming box C experienced a frictional force of 15 N, how much force does box C exert on box B? - Newton's third law of motion states that When two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. When solving this question, we must understand that forces exerted by box B and ¢ to one another are equal and opposite. - Given: mA: 8.00kg, mB=6.00kg, mC=12.0kg, a=2.82 m/s?, F,=-15N - required: FConB=? - solution: FnetB on c= mC x a = FBonC + Ff FBonc = mCa-F= 12.0kg x 2.82 m/s^2- (-15N) = 48.8N Because of Newton's third law of motion, FConB=-FBonC=-48.8N or 48.8N [left]

4. What is the effect of kinetic friction on an object? - A 5.0 kg box is being pulled with a force of 20 N and is sliding with an acceleration of 2.0 m/s. Find the coefficient of kInetic friction. - Solving this question requires understanding the net force is the sum of the applied force and the kinetic friction. - Given: m= 5.0 kg, Fapp=20N, a=2.0m/s^2 - Required: u=? - Solution: Fnet = Fapp + Ff ma=Fapp - umg M = (Fapp - ma) / mg = (20N - 5kgx2m/s^2) / 5kg×9.81m/s^2= 0.20

5. By adding vector components using numerical data and/or graphical data, what is the resultant force on an object? -A block of 80kg is supported by two cables with tension T1 and T2. Calculate T1 and T2. -This Is an example of using vector components of different forces acting on an object, to find the resultant force. - In the vertical plane. two forces are acting: the weight of the block acts downward while the tension of the cables pulls the block upward. - Since the forces in the vertical plane are balanced, T1sin60° + T2sin30° = 800N - In the horizontal plane too, the forces are balanced. So T1cos60°=T2cos30°, T1=square root3xT2 - square root3 x T2 x square root3/2 + T2 × ½ = 800 N, T2(3/2+½)=800N, T2= 800N/(3/2+½) = 400N - T1=square root3 x T2 = square root3 x 400N = 693N

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6. What is the difference in weight of an object on different planets? -Michael lands on a new planet, where the acceleration due to gravity is 1.56m/s2. If he weighed 196.2N on Earth what would his weight be on the new planet? Compare his weight on the planet with his weight on Earth - This question is an example of how weight varies from place to place in the universe depending on the gravitational acceleration. - m= 196.2N / 9.81m/s^2 = 20kg - W = 20kg x 1.56m/s^2 = 31.2N - He weighs 165 N less on the new planet than on Earth.

7. Using Newton’s law of universal gravitation how does the gravitational constant relate to the local value of acceleration due to gravity? How can you describe that relationship mathematically? -If a person of 55 kg mass stands on the surface of the moon, calculate the . The mass of the moon is 7.4 × 1022 kg, and the radius of the moon is 1.74 x 106 m. - The force of gravity anywhere is directly proportional to the product of the masses of two objects, and indirectly proportional to the square of the distance between their centers. The proportionality constant is universal anywhere, so we use it to calculate the force of gravity. Therefore, Newton's law of universal gravitation states that Fg= Gm1m2/r^2. since Fg=m1g, m1g= Gm1m2/r^2. From this, we derive g=Gm2/r^2 - So, g= (6.67 x 10^-11 Nm^2 /kg^2x 7.4 x 10^22 kg) / (1.74 x 10^6 m)^2 = 1.6 m/s^2

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