Physics Preparatory Course 2021 Part 2 PDF

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Physics Preparatory Course Lesson 4: Mechanics – 1D Kinematics, Momentum, Newton’s LawsDr Ho Shen Yong Lecturer, School of Physical and Mathematical Sciences Nanyang Technological University9/12 Jul 2021Even if you're on the right track, you'll get run over if you just sit there. Will RogersKey thin...

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B (L4 – L9) Physics Preparatory Course Lesson 4: Mechanics – 1D Kinematics, Momentum, Newton’s Laws Dr Ho Shen Yong Lecturer, School of Physical and Mathematical Sciences Nanyang Technological University 9/12 Jul 2021

Even if you're on the right track, you'll get run over if you just sit there. Will Rogers

Key things to committed to long term memory

1

∘ Motion in one dimension Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

Simple consideration 1 (Mazur)

You need to drive to a grocery store that is 2.0 km west of your house on the same street on which you live. There are three traffic lights between your house and the store, and on your trip you reach all three of them just as they change to red. While you are moving, your average speed is 30 km/h, but you have to slow down, stop, wait and move off at each traffic light and you the total time spent is 1 min at each traffic light. (You can ignore the size of the traffic junction for your calculations.) (a) (b) (c) (d)

How long does it take you to reach the store? What is your average velocity for the trip? What is your average speed? Do a sketch of how the speed of your car changes with time.

2

∘ Motion in one dimension Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

Simple consideration 2 (Mazur)

A helicopter pilot at an airport is told to fly a distance d at speed  directly east to pick up a stranded hiker. However, he sees no hiker when he arrives. Radioing the control tower, he learns that the hiker is actually a distance d west of the airport. The pilot turns around and heads straight for the hiker at a speed that is 50% faster than his speed during the initial segment of this wrong-way trip. (a) Calculate the total time. (b) What is the helicopter’s average velocity for the trip? (c) Do a sketch of how the velocity of your helicopter changes with time.

3

∘ Motion in one dimension Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

Simple consideration 3 (Mazur)

Six children—call them A, B, C, D, E, and F—are running up and down the street playing tag. Their positions as a function of time are plotted in the figure. The street runs east-west, and the positive x direction is eastward.

Use the figure above to answer these questions: (a) In which direction is each child moving? (b) Which, if any, of the children are running at constant velocity? Of those who are, is the x component of their velocity positive or negative? (c) Which, if any, of the children are not moving at constant velocity? Are they speeding up or slowing down? (d) Which child has the highest average speed? The lowest average speed? (e) Does any child pass another child during the time interval shown in the graph?.

4

∘ Motion in one dimension Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

𝑡 = 0𝑠

1𝑠

2𝑠

3𝑠

4𝑠

5𝑠

The position of a particle at regular time intervals is as shown in the diagram above. Describe the motion of the particle.

𝑡 = 0𝑠 1𝑠

2𝑠

3𝑠

4𝑠

The position of a particle at regular time intervals is as shown in the diagram above. Describe the motion of the particle.

𝑡 = 0𝑠 1𝑠

2𝑠

3𝑠

4𝑠

5𝑠

6𝑠

The position of a particle at regular time intervals is as shown in the diagram above. Describe the motion of the particle.

5

http://www.algodoo.com/

6

∘

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 Acceleration = −12.0 × 10−2 𝑚

A car moves off from rest from a traffic junction and the speed (velocity) increases uniformly. After 5 s, the speed of the car is 10 m/s. What is the rate of increase of the speed, (velocity) i.e. acceleration? Do a sketch using the axes below to illustrate how the velocity of the bus changes with time. If the car continues to accelerate uniformly, what is the velocity 3 s later? 𝑣

𝑡

7

∘ Acceleration Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

A motorcycle moves past at 10 m/s when the car just started to move. If the motorcycle has the same acceleration as the car, what will its speed be after 8 s? Sketch on the same axes, how the velocity of the motorcycle varies with time. How much further did the car travel? 𝑣

𝑡

8

∘ Acceleration Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

If acceleration 𝑎 is constant, we can write 𝑣𝑓 − 𝑣𝑖 𝑎= 𝑡 where 𝑣𝑓 is the final velocity and 𝑣𝑖 is the initial velocity and 𝑡 is the time taken for the change. We can sketch a graph of velocity against time. In the previous example, the velocity time graph for the car looks like the graph below. Vel. / m/s 𝑣𝑓

𝑣𝑖 0

𝑡 time / s

From the expression 𝑎=

𝑣𝑓 − 𝑣𝑖 𝑡

and the velocity time graph for constant acceleration, we can deduce some more information. 𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡 Displacement: 1

𝑥 = 𝑣𝑖 + 𝑣𝑓 𝑡 2 Displacement: 1

𝑥 = 𝑣𝑖 𝑡 + 𝑎𝑡 2 2 “No-time” equation: 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎𝑥

9

∘ Acceleration Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

Speeding up and slowing down

EXERCISE 1.

[G2.20] A sports car accelerates from rest to 95 km/h in 4.5 s. What is its average acceleration in m/s2 ?

2.

[G2.21] At highway speeds, a particular automobile is capable of an

acceleration of about 1.8 m/s2. At this rate, how long does it take to accelerate from 80 km/h to 110 km/h?

10

∘

3.

Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 10 down 𝑚 [G2.30] A car×slows from 25 m/s to rest in a distance of 85 m. What was its acceleration, assumed constant?

4.

[G2.32] A light plane must reach a speed of 32 m/s for takeoff. How long

a runway is needed if the (constant) acceleration is 3.0 m/s2?

5.

[G2.35] A world-class sprinter can reach a top speed (of about 11.5 m/s) in the first 15.0 m of a race. What is the average acceleration of this sprinter and how long does it take her to reach that speed?

11

−6 ∘

∘

∘

=unmarked 𝛼𝑙𝑜 Δ𝑇 =−2 12 ×car 10 traveling 𝐶 200𝑚 20 𝐶 (−30) 𝐶 6. [G2.44]Δ𝑙 a constant 95−km/h is passed by a =An −12.0 × 10 police 𝑚 speeder traveling 135 km/h. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car’s acceleration is 2.00 m/s2. How much time passes before the police car overtakes the speeder (assumed moving at constant speed)?

7. [G2.47] Mary and Sally are in a foot race. When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally thinks she has an easy win and so, during the remaining portion of the race, slows down at a constant rate of 0.50 m/s2 to the finish line. What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?

12

8.

∘

Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

13

Gradient, handling non∘ constant rate of change

Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

Previously, we have dealt with constant acceleration where the 𝑣 − 𝑡 graph is a straight line. When we have a curve, we have to talk about the gradient at a specific point. Graphically, we can sketch the tangent (line) at this point and then compute the gradient of this tangent. So essentially we finding the rate of change of 𝑥 in the limit when Δ𝑡 → 0 or writing in formal mathematics

Giancoli Fig 2.12

Δ𝑥 𝑑𝑥 = Δ𝑡→0 Δ𝑡 𝑑𝑡

𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = lim

So we see that at P1 and P2, 𝑥 is increasing with 𝑡 so the gradient is positive. In fact, the gradient is increasing from P1 to P2 before it levels off to zero gradient at P3. The value of 𝑥 is momentarily constant here and is a maximum for 𝑥 . After that, the gradient is negative and the value of 𝑥 is decreasing with time 𝑡 .

Giancoli Fig 2.10: Average velocity (from 𝑥 − 𝑡 )

Giancoli Fig 2.12: Instantaneous velocity

14

Instantaneous /∘ average velocity

Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚

Giancoli Fig 2.10: Average velocity (from 𝑥 − 𝑡 )

Giancoli Fig 2.12: Instantaneous velocity

Δx 𝑑𝑥 = Δ𝑡→0 Δ𝑡 𝑑𝑡

𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑣 = lim

Example [G2.4] A rolling ball moves from to during the time from 𝑥1 = 3.4 𝑐𝑚 to 𝑥2 = 24.2 𝑐𝑚 and during the time from 𝑡1 = 3.0 𝑠 to 𝑡 = 5.1 𝑠. What is its average velocity?

15

∘

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10Calculus 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 1 = −12.0 × 10−2 𝑚

If we know the equation of the line, is there a simpler way to obtain the gradient, other can obtaining it graphically? We need to know how to compute the quantity Δ𝑥 𝑑𝑥 = Δ𝑡→0 Δ𝑡 𝑑𝑡

𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = lim

to be able to find a gradient at point on a curve. Consider a relation, 𝑥 = 5𝑡 2 and consider two nearby points 𝑡1 and 𝑡1 + 𝛿𝑡 where 𝛿𝑡 is very small (say 0.0000001 or even smaller). Δ𝑥 5 𝑡1 + 𝛿𝑡 2 − 5 𝑡1 = Δ𝑡 (𝑡1 + 𝛿𝑡) − 𝑡1

=

5 𝑡12 + 2𝑡1 𝛿𝑡 + (𝛿𝑡)2

At 𝑡 = 𝑡1 and making 𝛿𝑡 → 0,

𝛿𝑡

− 5 𝑡1

2

2

= 5(2𝑡1 + 𝛿𝑡)

Δ𝑥 = lim 5(2𝑡1 + 𝛿𝑡) = 5 2𝑡1 = 10𝑡1 𝛿𝑡→0 Δ𝑡→0 Δ𝑡

𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = lim

Let’s see if we can find any pattern for polynomial functions of the form 𝑥 = 𝑘𝑡 𝑛 where 𝑘 is a constant and 𝑛 is an integer (for the time being). 𝑥 = 𝑘𝑡 3 𝑘 𝑡 3 + 3𝑡 2 𝛿𝑡 + 3𝑡 𝛿𝑡 2 + 𝛿𝑡 𝑘 𝑡 + 𝛿𝑡 3 − 𝑘𝑡 3 Δ𝑥 = = 𝛿𝑡 (𝑡 + 𝛿𝑡) − 𝑡 Δ𝑡 2 2 = k 3𝑡 + 3𝑡𝛿𝑡 + 𝛿𝑡 𝑑𝑦 = lim k 3𝑡 2 + 3𝑡𝛿𝑡 + 𝛿𝑡 𝑑𝑡 𝛿𝑡→0

2

3

−𝑘 𝑡

3

= 𝑘 3𝑡 2 = 3𝑘𝑡 2

16

∘

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10Calculus 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 1 = −12.0to× 10−2 𝑚 We can generalize 𝑑𝑥

𝑥 = 𝑘𝑡 𝑛 ⇒ = 𝑘 𝑛𝑡 𝑛−1 𝑑𝑡 This finding is even applicable for when 𝑛 are negative integers, fractions and even irrational numbers such as √2 and 𝜋. (You can check these out after you learn how to do binomial expansion of non-positive integer exponents.) Exercise 𝑑𝑦 For 𝑦 = 7𝑥 3 − 4𝑥 + 2, compute and evaluate it at 𝑥 = 3.2. 𝑑𝑥

We will discuss about areas and integration later. Exercise The position of a small object is given by 𝑥 = 34 + 10𝑡 − 2𝑡 3 where 𝑡 is in seconds and 𝑥 in meters. (a) Plot 𝑥 as a function of 𝑡 from 𝑡 = 0 to 𝑡 = 3.0 𝑠. (b) Find the average velocity of the object between 0 and 3.0 s. (c) At what time between 0 and 3.0 s is the instantaneous velocity zero?

17

Instantaneous /∘ average velocity

Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, 𝑣𝑓 − 𝑣𝑖 Δv 𝑎ത = = Δ𝑡 𝑡𝑓 − 𝑡𝑖 𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, Δv 𝑑𝑣 𝑑 𝑑𝑥 𝑑2 𝑥 𝑎 = lim = = = 2 Δ𝑡→0 Δ𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡

An example of non-constant acceleration is the acceleration of a swinging pendulum. (Talk about trigonometric functions)

displacement

18

∘

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10Free 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 Falling −2 −12.0A×1 10 𝑚 and a 1 g object are released from the same Pause to = Ponder: kg object

height at the same time in a vacuum. Which object will hit the ground first? a. b. c.

1 kg 1g At the same time

19

∘

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10Free 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 Falling = −12.0 × 10−2 𝑚

https://www.youtube.com/watch?v=E43-CfukEgs

Giancoli Figure 2.27

In the absence of air resistance, all objects fall with the same acceleration, although this may be tricky to tell by testing in an environment where there is air resistance. We will prove this fact after we have discussed Newton’s second law. Exercise: [For simplicity, take 𝑔 = 10 𝑚/𝑠 2 ] An object at rest is released from a height of 2.0 𝑚. Sketch a graph to show how the velocity and acceleration of the object changes with time.

20

Free Falling∘– 3 Scenarios

Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = simplicity −12.0 ×take 10 𝑔 = 𝑚10 𝑚 𝑠 −2) Free fall (For a. An object at rest is released from a height of 2.0 𝑚. What is the velocity of the object when it hits the ground? How much time does it take to hit the ground?

b. Now the object is thrown vertically downwards at 10.0 𝑚 𝑠 −1 from a height of 8.0 𝑚. What is the velocity when it hits the ground? How much time does it take?

[email protected]

21

Free Falling∘– 3 Scenarios

Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = simplicity −12.0 ×take 10 𝑔 = 𝑚10 𝑚 𝑠 −2) Free fall (For c. Now the object is thrown vertically upwards at 10.0 𝑚 𝑠 −1 from a height of 8.0 𝑚. What is the velocity when it hits the ground? How much time does it take?

Sketch the velocity-time graph for each case. 𝑣

𝑣

𝑣

𝑡

𝑡

𝑡

[email protected]

22

∘

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10Free 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 Falling = −12.0 × 10−2 𝑚 An object is thrown vertically upwards with an initial velocity of 5.0 𝑚 𝑠−1 . Sketch the

velocity-time graph, acceleration-time graph and displacement-time graph. Take g = 10 ms-2. Take downwards as +ve. 𝑎

𝑡

𝑣

𝑡

𝑠

𝑡

[email protected]

23

∘

1.

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10EXERCISE 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10−2 𝑚

24

∘

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10EXERCISE 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 10−2 𝑚vertically upward with a speed of 24.0 m/s. (a) How 2. [G2.60]=A−12.0 stone × is thrown

fast is it moving when it reaches a height of 13.0 m? (b) How much time is required to reach this height? (c) Why are there two answers to (b)?

3. [G2.65] A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4 s later. If the speed of sound is 340 m/s, how high is the cliff?

25

∘

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10EXERCISE 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 −2 −12.0 × 10 𝑚 4. [G2.89]=Agent Bond is standing on a bridge, 13 m above the road below, and his

pursuers are getting too close for comfort. He spots a flatbed truck approaching at 25 m/s, which he measures by knowing that the telephone poles the truck is passing are 25 m apart in this country. The bed of the truck is 1.5 m above the road, and Bond quickly calculates how many poles away the truck should be when he jumps down from the bridge onto the truck, making his getaway. How many poles is it?

26

∘

−6 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10EXERCISE 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 −2 −12.0 × 10street 𝑚 pattern shown in the figure. Each intersection has a 5. [G2.78]=Consider the

traffic signal, and the speed limit is 50 km/h. Suppose you are driving from the west at the speed limit. When you are 10.0 m from the first intersection, all the lights turn green. The lights are green for 13.0 s each. (a) Calculate the time needed to reach the third stoplight. Can you make it through all three lights without stopping? (b) Another car was stopped at the first light when all the lights turned green. It can accelerate at the rate of 2.00 m/s2 to the speed limit. Can the second car make it through all three lights without stopping? By how many seconds would it make it or not?

27

Physics Preparatory Course Lesson 5: Mechanics – 1D Kinematics, Momentum, Newton’s Laws Dr Ho Shen Yong Lecturer, School of Physical and Mathematical Sciences Nanyang Technological University 14 Jul 2021

Failure is only the opportunity to begin again more intelligently. Henry Ford

Key things to committed to long term memory

28

∘

∘ ∘ Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10−6Forces 𝐶 200𝑚 (Knight)20 𝐶 − (−30) 𝐶 = −12.0 × 10−2 𝑚

•

A force is a push or a pull.

•

A force acts on an object.

•

Every force has an agent, something that acts or pushes or pulls.

•

Ԧ A force is a vector. The general symbol for a force is the vector symbol 𝐹. The size or strength of such a force is its magnitude F.

•

Contact forces are forces that act on an object by touching it at a point of contact.

•

Long-range forces are forces that act on an object without physical contact.

Contact Forces

Long Range Forces

29

∘

∘ ∘ Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 = 12 × 10−6Forces 𝐶 200𝑚 (Knight)20 𝐶 − (−30) 𝐶 = −12.0 × 10−2 𝑚

30

Weight

∘ Types (Knight)20∘ 𝐶 − (−30)∘ 𝐶 Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10of−6Forces 𝐶 200𝑚 = −12.0 × 10 𝑚

•

The gravitational pull of the earth on an object on or near the surface of the earth is called weight.

•

The agent for the weight forces is the entire earth pulling on an object.

•

An object’s weight vector always points vertically downward, no matter how the object is moving.

Upthrust / Bouyant Force •

Archimedes' Principle: Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

1.

Volume of object submerged in the fluid=Volume of fluid displaced;

2.

Knowing the displaced volume Vsubmerged and density rfluid of the fluid, we can compute the weight of fluid which gives the upthrust: 𝑈𝑝𝑡ℎ𝑟𝑢𝑠𝑡 = (𝜌𝑓𝑙𝑢𝑖𝑑 𝑉𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 ) 𝑔

31

∘

Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚 • Springs come in in many forms. When deflected, they push or pull with a Spring Force

spring force.

• •

The magnitude of the force is dependent on the extension or compression of the spring. If the magnitude of the spring force 𝐹𝑠𝑝 is directly proportional to extension / compression 𝑥, we say that the spring obeys Hooke’s law: 𝐹𝑠𝑝 = 𝑘𝑥

where 𝑘 is known as the spring constant – the stiffer the spring, the larger will be the value of 𝑘 .

Sometimes, we write the vector version of the equation: 𝐹Ԧ𝑠𝑝 = −𝑘𝑥Ԧ

The negative sign :

32

∘

Δ𝑙 = 𝛼𝑙𝑜 Δ𝑇 =−212 × 10−6 𝐶 200𝑚 20∘ 𝐶 − (−30)∘ 𝐶 = −12.0 × 10 𝑚 • When a string or rope or wire pulls on an object, it exerts a contact force that Tension Force

we call the tension force.

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The direction of the tension force is always in the direction of the string or rope.

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The direction of the tension is always pointing away from the object.

Exercise A ball rolls down an incline and off a horizontal ramp. Ignoring air resistance, what force or forces act on the ball as it moves through the air just after leaving the horizontal ramp?

A. B. C. D. E.

The weight of the ball acting vertically down. A horizontal force that maintains the motion. A force whose direction changes as the direction of motion changes. The weight of the ball and a horizontal force. The weight of the ball and a force in the direction of...