Forces Quest Answer key PDF

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jonnalagadda (saj2436) – Forces – ayida – (4355-3) This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An empty 200 kg elevator accelerates upward at 1.3 m/s2. The acceleration of gravity is 9.8 m/s2 . What is the tension in the cable that lifts the elevator cab?

1

1.

2.

Correct answer: 2220 N. Explanation: 3.

According to Newton’s second law,

4.

Fnet = T − m g = m a T = m(g + a) = (200 kg)(9.8 m/s2 + 1.3 m/s2) = 2220 N

002 10.0 points A man stands in an elevator in the university’s administration building and is accelerating upwards. (During peak hours, this does not happen very often.)

5.

correct 6.

Choose the correct free body diagram for the man, where Fi,j is the force on the object i, from the object j.

Explanation: Only the forces acting directly on the man

jonnalagadda (saj2436) – Forces – ayida – (4355-3) are to be in the free body diagram. Therefore, the force from the cable should be omitted, while those from gravity and from the floor’s normal force should be included.

2

reads Su . When it is moving downward with the same constant speed v , the scale reads Sd .

003 (part 1 of 2) 10.0 points An elevator accelerates upward at 1.2 m/s2 . The acceleration of gravity is 9.8 m/s2 . What is the upward force exerted by the floor of the elevator on a(n) 92 kg passenger? Correct answer: 1012 N. Explanation:

Scale When the elevator is accelerating upward, Fnet = m a = N − m g N = m a + m g.

Choose the correct relationship among the three scale readings. 1. Su > S0 > Sd 2. Su = Sd , Su > S0

004 (part 2 of 2) 10.0 points If the same elevator accelerates downwards with an acceleration of 1.2 m/s2 , what is the upward force exerted by the elevator floor on the passenger? Correct answer: 791.2 N. Explanation:

When the elevator is accelerating downward,

3. Su < S0 < Sd 4. S0 = Su = Sd correct 5. Su = Sd , Su < S0 Explanation: Consider the free body diagram for each the case where the elevator is accelerating down (left) and up (right). The man is represented as a sphere and the scale reading is represented as S . Su a Sd a

Fnet = m a = m g − N2 N2 = m g − m a. mg

mg

005 (part 1 of 2) 10.0 points Consider a man standing on a scale in an elevator. When the elevator stays at rest, the scale reads S0 . When the elevator is moving upward with a constant speed v, the scale

Note: The elevator is not accelerating, but moving upward or downward with constant speed! We consider the forces acting on the man. Let up be positive. Then the gravitational

jonnalagadda (saj2436) – Forces – ayida – (4355-3)

3

~ = −m g ˆ. The only force on the man is W other force acting on the man is the normal force ~S from the scale. By Newton’s third law, the force on the scale exerted by the man (i.e., the scale reading) is equal in magnitude but opposite in direction to the ~S vector. However for all three cases, the acceleration is zero. Thus by Newton’s second law, for all three cases

Looking at the left-hand free body diagram and using Newton’s second law, we have

S − mg = 0 S = mg ,

007 (part 1 of 2) 10.0 points A light, inextensible cord passes over a light, frictionless pulley with a radius of 15 cm. It has a(n) 13 kg mass on the left and a(n) 2.8 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 2 m apart. The acceleration of gravity is 9.8 m/s2 .

or S0 = Su = Sd . 006 (part 2 of 2) 10.0 points Again, when the elevator stays at rest, the scale reads S0. Suppose the elevator now accelerates downward at a constant rate of 1 g. 5 What is the new scale reading? g is the magnitude of the gravitational acceleration. 1. Sd =

6 S0 5

1 Sscale − m g = − m g 5 4 4 Sscale = m g = S0 . 5 5

15 cm ω

13 kg

2. Sd = S0 3. Sd =

7 S0 5

4. The scale reading is decreasing with time. 5. Sd = 2 S0 6. The scale reading is increasing with time. 3 S0 5 4 8. Sd = S0 correct 5 1 9. Sd = S0 5 2 10. Sd = S0 5 Explanation: 7. Sd =

2m 2.8 kg At what rate are the two masses accelerating when they pass each other? Correct answer: 6.32658 m/s2 . Explanation:

Let : R = 15 cm , m1 = 2.8 kg , m2 = 13 kg , h = 2 m , and v = ωR.

Consider the free body diagrams

a

1.9 kg

m1 g

2.8 kg

m2 g

a

13 kg

Since the larger mass will move down and the smaller mass up, we can take motion downward as positive for m2 and motion upward as positive for m1 . Apply Newton’s second law to m1 and m2 respectively and then combine the results: For mass 1: X F 1 : T − m1 g = m1 a (1) For mass 2: X F2 :

4

A mass of 1.9 kg lies on a frictionless table, pulled by another mass of 4.4 kg under the influence of Earth’s gravity. The acceleration of gravity is 9.8 m/s2 .

T

T

jonnalagadda (saj2436) – Forces – ayida – (4355-3)

4.4 kg What is the magnitude of the acceleration a of the two masses? Correct answer: 6.84444 m/s2 . Explanation: Given :

m2 g − T = m2 a

m1 = 1.9 kg and m2 = 4.4 kg .

(2)

We can add Eqs. (1) and (2) above and obtain:

a

m2 g − m1 g = m1 a + m2 a m2 − m1 g m1 + m2 13 kg − 2.8 kg = (9.8 m/s2 ) 13 kg + 2.8 kg = 6.32658 m/s2 .

a=

T T

m1 N

m1 g

a

m2 g

Let the direction of acceleration as indicated in the figure be positive. The net force on the system is simply the weight of m2 . Fnet = m2 g = 43.12 N .

008 (part 2 of 2) 10.0 points What is the tension in the cord when they pass each other?

From Newton’s second law,

Correct answer: 45.1544 N.

Solving for a,

Explanation:

m2

Fnet = m2 g = (m1 + m2 ) a .

m2 g m1 + m2 4.4 kg = 1.9 kg + 4.4 kg = 6.84444 m/s2 .

a= T = m1 (g + a) = (2.8 kg) (9.8 m/s2 + 6.32658 m/s2 ) = 45.1544 N . 009

10.0 points

010 (part 1 of 2) 10.0 points A pulley is massless and frictionless.

The

jonnalagadda (saj2436) – Forces – ayida – (4355-3) masses 1 kg, 2 kg, and 10 kg are suspended as in the figure.

22 cm ω T2

T 1 − m1 g = m1 a , for the upper left-hand mass m2 the acceleration is up and

3.7 m

T1

acceleration of the system will be down to the right (m3 > m1 + m2 ), and each mass in the system accelerates at the same rate (the string does not stretch). Let this acceleration rate be a and the tension over the pulley be T ≡ T2 = T3 . For the lower left-hand mass m1 the acceleration is up and

T3

2 kg

5

10 kg

T − T 1 − m2 g = m2 a ,

1 kg What is the tension T1 in the string between the two blocks on the left-hand side of the pulley? The acceleration of gravity is 9.8 m/s2 .

and for the right-hand mass m3 the acceleration is down and −T + m3 g = m3 a . Adding,

Correct answer: 15.0769 N. (m3 − m1 − m2 ) g = (m1 + m2 + m3 ) a .

Explanation: Let : R = 22 cm , m1 = 1 kg , m2 = 2 kg , m3 = 10 kg , and h = 3.7 m .

m3 − m1 − m2 g m1 + m2 + m3 10 kg − 1 kg − 2 kg (9.8 m/s2 ) = 1 kg + 2 kg + 10 kg = 5.27692 m/s2 .

a=

T1

10 kg

a

T3

The tension in the string between block m1 and block m2 (on the left-hand side of the pulley) can be determined from the equation for m1 :

m3 g

T2 2 kg m2 g

1 kg m1 g

a

T1

Consider the free body diagrams

For each mass in the system ~ Fnet = m~a . Since the string changes direction around the pulley, the forces due to the tensions T2 and T3 are in the same direction (up). The

T1 = m1 (a + g) = (1 kg) (5.27692 m/s2 + 9.8 m/s2 ) = 15.0769 N .

011 (part 2 of 2) 10.0 points What is the magnitude of the acceleration of the lower left-hand block? Correct answer: 5.27692 m/s2 . Explanation:

jonnalagadda (saj2436) – Forces – ayida – (4355-3) The acceleration is the same for every mass, since the string is inextensible.

T

α

m

012 10.0 points The block of mass 4.78893 kg has an acceleration of 4.6 m/s2 as shown.

µs Which choice best describes the free body diagram in the vertical direction for this situation?

F 20◦

a

6

4.78893 kg

1. N = m g + T sin α What is the magnitude of F ? Assume the acceleration due to gravity is 9.8 m/s2 and the surface is frictionless.

2. N = m g + T cos α

Correct answer: 23.4429 N.

4. N = m g − T sin α correct

Explanation:

Let :

5. N = m g − T cos α Explanation:

F = 23.4429 N , α = 20◦ , m = 4.78893 kg , g = 9.8 m/s2 , F

ng

T stri and

α

m mg

3. N = m g

N

Fnet = m a = F cos α ma F = cos α (4.78893 kg) (4.6 m/s2 ) = cos 20◦ = 23.4429 N .

m

µs N mg

N

Because the string is pulling partially up, the force of the table on the book (the normal force) is less than if the string were not there. X Fy : N + T sin α = m g N = m g − T sin α .

014 (part 2 of 2) 10.0 points Which forces must change in order for the book to start moving? 1. normal, friction, and tension in the string correct 2. None of these

013 (part 1 of 2) 10.0 points A string is tied to a book and pulled at an angle α as shown in the diagram. The book remains in contact with the table and does not move.

α

3. normal 4. tension in the string 5. normal and gravitational

jonnalagadda (saj2436) – Forces – ayida – (4355-3)

7. friction

F2

6. friction, tension in the string, and gravitational F1

Explanation: The tension T must increase, which increases T sin α and decreases the normal force N. The friction forces depends directly on the normal force N (the book will slide more easily if the normal force decreases), so the friction force also decreases.

68

N

015 10.0 points The horizontal surface on which the block of mass 5.5 kg slides is frictionless. The force of 34 N acts on the block in a horizontal direction and the force of 68 N acts on the block at an angle as shown below.

34 N

60◦

5.5 kg

What is the magnitude of the resulting acceleration of the block? The acceleration of gravity is 9.8 m/s2 . Correct answer: 0 m/s2 . Explanation:

Let :

F1 = 34 N , F2 = 68 N , α = 60◦ , m = 5.5 kg , and g = 9.8 m/s2 .

α

m

8. gravitational 9. friction and tension in the string

7

N

mg

The force F2 has a horizontal component F2 cos α acting to the left, and the force F1 acts to the right, so Fnet = m a = F2 cos α − F1 ◦ = F2 cos 60 − F1   1 − F1 F2 2 a= m   1 − 34 N 68 N 2 =0. = 5.5 kg 016 10.0 points A 1300 kg car moves along a horizontal road at speed v0 = 27 m/s. The road is wet, so the static friction coefficient between the tires and the road is only µs = 0.16 and the kinetic friction coefficient is even lower, µk = 0.112. The acceleration of gravity is 9.8 m/s2 . What is the shortest possible stopping distance for the car under such conditions? Use g = 9.8 m/s2 and neglect the reaction time of the driver. Correct answer: 232.462 m. Explanation: The force stopping the car is the friction force Ff between the tires and the road. If the car skids, the friction force is governed by the kinetic friction law, Ff = µk N where N is the normal force between the car and the road. If the car does does not skid, the friction is static and the friction force can be anything up to a maximal static friction Ffmax = µs N . Given µs > µk , it follows that static friction at its strongest is stronger than kinetic friction: The strongest possible friction force is Ffmax = µs N

jonnalagadda (saj2436) – Forces – ayida – (4355-3) which obtains only when the car does not skid. The normal force between the car and the road follows from Newton’s Laws for the vertical direction of motion: Since the car moves in the horizontal direction only, ay = 0 and hence Fynet = N − mg = 0. Consequently, N = mg ,

Ffmax = µs mg, and since there are no horizontal forces other than friction, the acceleration is limited to |ax | ≤ amax =

1 max F = µs g. m f

Now consider the kinematics of the car’s deceleration to stop. Clearly, the shortest stopping distance obtains for the maximally negative ax , hence ax = −amax = −µs g = const. At constant deceleration, the stopping time of the car is v0 , ts = |ax | which gives us the stopping distance Xs = v0 ts −

|ax | 2 v2 ts = 0 . 2 2|ax |

For the problem at hand, |ax | = µs g, hence the stopping distance Xs =

v02 = 232.462 m. 2µs g

Note: The answer does not depend on the car’s mass m. 017 (part 1 of 3) 10.0 points You are driving at the speed of 28.4 m/s (63.5426 mph) when suddenly the car in front of you (previously traveling at the same speed) brakes. Considering an average human reaction, you press the brakes 0.501 s later. Assume that the brakes on both cars are fully engaged and that the coefficient of friction is 0.922 between both cars and the road.

8

The acceleration of gravity is 9.8 m/s2 . Calculate the magnitude of the acceleration of the car in front of you when it brakes. Correct answer: 9.0356 m/s2 . Explanation: The only force on the car horizontally when braking is the force of friction, Ff . Thus Fnet = ma = −Ff = −µN = −µmg since the braking force is a deceleration. Thus ma = −µmg a = −µg = −9.0356 m/s2 . 018 (part 2 of 3) 10.0 points Calculate the braking distance for the car in front of you. Correct answer: 44.6323 m. Explanation: For constant acceleration a, v 2 − v02 = 2a (x − x0)

.

The final velocity vf = 0, and for dbr = x−x0 , −v02 = 2 a dbr v2 v2 dbr = − 0 = 0 2µg 2a = 44.6323 m 019 (part 3 of 3) 10.0 points Find the minimum safe distance at which you can follow the car in front of you and avoid hitting it (in the case of emergency braking described here). Correct answer: 14.2284 m. Explanation: The car in front of you moves a total distance of dbr , which also is the distance you travel while braking. You also traveled a distance dreact = v0 ∆t because of your reaction time ∆t. Thus the minimum safe distance is

jonnalagadda (saj2436) – Forces – ayida – (4355-3) 173 N 25.7◦

the distance you travel during your reaction time: dsaf e = v0 ∆t = 14.2284 m 020 10.0 points A horizontal force of 100 N is used to pull a crate, which weighs 500 N, at constant velocity across a horizontal floor. What is the coefficient of kinetic friction? 1. 20

Fk

4. 2

40.7 kg Fg

µ = 0.17 Fn

Note: Figure is not drawn to scale. Basic Concepts: ~ Fnet = ΣF~ = m~a Parallel to the floor, Fx = F cos θ

2. 0.2 correct 3. 0.5

9

Fk = µk Fn Fx,net = ΣFx = Fapplied,x − Fk = max

Perpendicular to the floor,

Fy = F sin θ

5. 0.05 Explanation: Draw the free body diagram to identify the forces acting on the crate. There is a force F =100 N acting to the right, a normal force N = m g = 500 N acting upward, and a frictional force Ffriction = µk N acting to the left (opposing the motion of the crate). Since the crate is being pulled at constant velocity, the acceleration is zero. The sum of the of forces in the horizontal direction gives F − Ffriction = F − µk N = 0 F 1 100 N µk = = . = 5 500 N N 021 (part 1 of 2) 10.0 points A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 173 N at an angle of 25.7◦ above the horizontal. The box has a mass of 40.7 kg, and µk between the box and the floor is 0.17. The acceleration of gravity is 9.81 m/s2 . Find the acceleration of the box. Correct answer: 2.47579 m/s2 . Explanation:

Fy,net = ΣFy = Fn + Fapplied,y − Fg = 0 Given: Fapplied = 173 N at 25.7◦ above the horizontal m = 40.7 kg µk = 0.17 g = 9.81 m/s2 Solution: Perpendicular to the floor, Fapplied,y = Fapplied sin θ = (173 N) sin 25.7◦ = 75.023 N Fn = Fg − Fapplied,y = mg − Fapplied,y

= (40.7 kg)(9.81 m/s2) − 75.023 N = 324.244 N

Parallel to the floor, Fapplied,x = Fapplied cos θ = (173 N) cos 25.7◦ = 155.886 N

jonnalagadda (saj2436) – Forces – ayida – (4355-3) Thus the horizontal acceleration is

Perpendicular to the ramp:

Fapplied,x − µk Fn m 155.886 N − (0.17)(324.244 N) = 40.7 kg = 2.47579 m/s2

ax =

Fg,y = mg cos θr = (40.7 kg)(9.81 m/s2 ) cos 14.5◦ = 386.549 N Fapplied,y = Fapplied sin θ = (173 N) sin 25.7◦ = 75.023 N

along the floor. 022 (part 2 of 2) 10.0 points Now the student moves the box up a ramp (with the same coefficient of friction) inclined at 14.5◦ with the horizontal. b) If the box starts from rest at the bottom of the ramp and is pulled at an angle of 25.7◦ with respect to the incline and with the same 173 N force, what is the acceleration up the ramp?

Fn = Fg,y − Fapplied,y = 386.549 N − 75.023 N = 311.526 N Parallel to the ramp: Fg,x = mg sin θr

Correct answer: 0.0726871 m/s2.

= (40.7 kg)(9.81 m/s2 ) sin 14.5◦ = 99.9685 N

Explanation:

173

F g ,x Fk 14.5◦

N ◦ 7 25.

0 µ= F g ,y

Fn

Fg,x = Fg sin θr = mg sin θr Fk = µk Fn Fx,net = Σ Fx = Fapplied,x − Fk − Fg,x = max

Perpendicular to the ramp,

Fg,y = Fg cos θr = mg cos θr Fy,net = Σ Fy = Fn + Fapplied,y − Fg,y = 0 θr = 14.5◦ Solution:

Fapplied,x = Fapplied cos θ = (173 N) cos 25.7◦ = 155.886 N

.17

Note: Figure is not drawn to scale. Basic Concepts: Parallel to the ramp,

Given:

10

Fk = µk Fn = (0.17)(311.526 N) = 52.9595 N and the acceleration is Fapplied,x − Fk − Fg,x m 155.886 N − 52.9595 N − 99.9685 N = 40.7 kg = 0.0726871 m/s2

ax =

up the ramp. 023 10.0 points Two blocks connected by a string are pulled across a rough horizontal surface by a force applied to one of the blocks, as shown. The acceleration of gravity is 9.8 m/s2 .

jonnalagadda (saj2436) – Forces – ayida – (4355-3) F

6 kg

T

◦

42

6 kg µ = 0.22

If each block has an acceleration of 3.1 m/s2 to the right, what is the magnitude of the applied force?

11

024 (part 1 of 3) 10.0 points A hockey puck is hit on a frozen lake and starts moving with a speed of 14.7 m/s. Four seconds later, its speed is 7 m/s. What is its average acceleration? The acceleration of gravity is 9.8 m/s2 . Correct answer: −1.925 m/s2 . Explanation:

Correct answer: 70.8393 N. Explanation:

Let : Given : M1 = 6 kg , M2 = 6 kg , θ = 42◦ , a = 3.1 m/s2 , µ = 0.22 . For the mass M1,

N1 = W1,

and

Horizontally,

so

Fnet = M1 a = T − µ M1 g .

vi = 14.7 m/s , ∆t...