Gr 11 Physics - Dynamics Review
1. List and define the four fundamental forces of nature.
i) Gravitational Force (Fg) – every object with mass has an attraction toevery other mass in the universe
ii) Electromagnetic Force – (Fe) – the force that charged particles have on each other
iii) Strong Nuclear Force – an attractive force that holds protons and neutrons in the nucleus of the atom acting in the 10-15 m range
iv) Weak Nuclear Force – the force exerted between all subatomic particlesenabling a conversion of one type of quark into another
2. If forces on an object are unbalanced what else is true about the situation?
Accelerated motion, ΣF=ma, a≠0,
3. State Newton’s Laws and give an example of each of them.
Law 1: Objects will remain at rest or in uniform straight line motion unless acted upon by a net force.Ie: person falls forward or backwards on a bus
Law 2: Acceleration is directly proportional to force and inversely proportional to mass.Ie: a full shopping cart is harder to turn than empty
Law 3: For every action there is an equal and opposite reaction but not on the same object. Ie: walking, you push back on the ground and the ground pushes you forward.
4. Explain what is meant by net force.
Net force is the vector sum of all forces on an object as opposed to a single force.
5. Give three examples of scalars and three examples for vectors.
Scalars: 24 points in basketball, height of 153 cm, 365 days in a year
Vectors: 150 m [N] of the flagpole, 320 N [down] weight of table, 2.5 m/s2 [S]
6. Add or subtract the following vectors by sketching. Display the addition process and the resultant.
a. + =c. +=
b. -x= d. +=
x =
7. Explain what is meant by collinear, orthogonal and non-orthogonal vector addition.
Collinear: along the same line either in the same or opposite direction
Orthogonal: at a right angle
Non-orthogonal: neither collinear or orthogonal
8. Adding vectors should be done ______to ______. (head to tail)
9. A person walks 12.5 km south and then 15.5 km east then 3.2 km north in 3.11 hours. What is the speed and velocity for the trip? Make sure to include a diagram.
Speed = distance/time = 31.2km/3.11h = 10.0 km/h
Velocity = displacement/timedisplacement:(12.5-3.2)2 + (15.5)2 = R2
R = 18.1 kmΘ = tan-1(9.3/15.5) R = 18.1 km [E31°S]
Velocity = 18.1 km [E31°S]/3.11h =
= 5.82 km/h [E31°S]
10. What is the weight of a 55 kg object on the surface of the Earth? (g=9.81 m/s2)
Fg=mg = 540 N
11. What is the weight of a 55 kg object on the surface of Mars? (g=3.30 m/s2)
Fg=mg = 180 N
12. Name the opposite vector to 25.0 N [N 27º E]. Opp: 25.0 N [S 27º W].
13. Resolve 25.0 N [N 27º E] into its respective components.
Ax = 25sin27 = 11.3 N [E]Ay = 25cos27= 22.3 N [N]
14. Scott hikes 12.45 km [N25.0ºW] then 23.81 km [S41ºE]. Using the component method, describe the vector that will get him back to where he began his hike.
Resolve, add collinear, add orthogonal
Ax = 12.45sin25 = 5.2616 km [W]Ay= 12.45cos25 = 11.2835 km [N]
Bx = 23.81sin41 = 15.6208 km [E]By = 23.81cos41 = 17.9696 km [S]
Ax + Bx = -12.45sin25 + 23.81sin41 = 10.359 km [E] (Cx)
Ay + By = 12.45cos25 – 23.81cos41 = 6.686 km [S] (Cy)
C2 = 10.3592 + 6.6862 C = 12.3 kmΘ = tan-1(6.686/10.359) Θ= 32.8°
C = 12.3 km [E33°S] but the vector back to his starting point would be the opposite vector, 12.3 km [W33°N]
*15. A plane can fly at 455 km/h with respect to the ground. The plane makes its heading directly north at this speed but an easterly crosswind of 90.0 km/h affects the flight direction. FYI: winds are named from the direction they come from. With respect to the ground, what is the resultant vector of this plane?
R = √ (4552 + 902)
R = 464 km/h
Θ = tan-1(90/455)
Θ = 11.2°
R = 464 km/h [N11.2°W]
**16. A plane needs to fly directly west. It can fly at 325 km/h with respect to the ground. If is known that there is a southwest wind at 85.0 km/h, what should the plane’s heading be if it is to maintain its due west heading?
so heading [W10.7
17. Draw the forces on an object that is being pulled at a constant velocity across a wooden floor with friction.
18. What is the mass of a cat that weighs 30.0 Newtons?
Fg=mg so m = 3.06 kg
19. What applied force is required to accelerate a 4.0 kg bowling ball at 5.0 m/s2 neglecting friction? ΣF=ma ΣF=2.0x101 N
Fnet= Fg + FN + FA
FnetH = FA = ma
FnetV= Fg + FN = 0
20. If the applied force required in qu 19 was 35 N, calculate the frictional force. What is the coefficient of friction?
Fnet= Fg + FN + FA + Ff
FnetH = FA + Ff= ma
FnetV= Fg + FN = 0
μK= Ff/FN
FN = -Fg = -mg = 49.05 N
FA + Ff= ma
Ff = ma - FA = 20-35 = -15 N
μK= 15 N/49.05 N = 0.306
21. A 78.2 kg box is pulled horizontally at a constant velocity across a warehouse floor with a 250 N force. What is the coefficient of friction?
Fnet= Fg + FN + FA + Ff
FnetH = FA + Ff= 0
FnetV= Fg + FN = 0
μK= Ff/FN
FN = -Fg = -mg = 767 N
FA + Ff= 0
Ff = 250 N
μK= 250 N/767 N
μK = 0.326
22. A pilot stops a plane of mass 45000 kg in a distance of 485 m. If the plane had a uniform acceleration of -8.00 m/s2, how fast was the plane moving before the braking began? What is the average force exerted on the plane through friction?
Fnet= Fg + FN + Fft =
FnetV = Fg + FN = 0a = -8.00 m/s2
FnetH = Ff = ma d = 485 m
Ff = 45000*8vi= ?
Ff = 360000 Nvf = 0
vi = √ (vf2 – 2ad)
a = 88.1 m/s
23. A force of -5.00 x 103 N is used to stop an 1800 kg car traveling at 30 m/s. What braking distance is needed to bring the car’s velocity to 0?
Fnet= Fg + FN + Fft = ?
FnetV = Fg + FN = 0a = -2.78 m/s2
FnetH = Ff = ma d = ?
a= -5.00 x 103 /1800vi = 30 m/s
a= - 2.78 m/s2vf = 0
d=(vf2 – vi2)/(2a)
d = 162 m