Download presentation

Presentation is loading. Please wait.

Published byGreta Pentecost Modified over 2 years ago

2
Part 3 /3 High School by SSL Technologies

4
Physics Ex-38 Question-1 A force of 12 N, acting 60 o from the horizontal, is applied to a 20 kg cart initially at rest resulting in a final velocity of 10 m/s. If the force of friction is 2 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 250 m). Click Was the cart at rest? a) No What was the applied force? b) 12 N [E 60 o N] What was the horizontal component of the applied force? c) 6 N right What was the frictional force? d) 2 N left Click What was the resultant force? e) 4 N right Click (While traveling from point-A to point-B) (F R = F A – f = 6 N – 2 N = 4 N) s = 250 m v i = 0v f = 10 m/s 20 kg f = 2 N A B F A = 12 N 60 o F H = 6 N “velocity not constant”

5
Physics Ex-38 Question-1 A force of 12 N, acting 60 o from the horizontal, is applied to a 20 kg cart initially at rest resulting in a final velocity of 10 m/s. If the force of friction is 2 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 250 m). Click s = 250 m v i = 0v f = 10 m/s 20 kg f = 2 N A B F A = 12 N 60 o F H = 6 N What was the acceleration of the cart? f) 0.2 m/s 2 What was the initial E K of the cart? g) 0 What was the final E K of the cart? h) 1 000 J How much work was done on the cart? i) 1000 J What becomes of the work done on the cart? j) It is transferred to the cart in the form of E K (faster speed). Click

6
Physics Ex-38 Question-1 A force of 12 N, acting 60 o from the horizontal, is applied to a 20 kg cart initially at rest resulting in a final velocity of 10 m/s. If the force of friction is 2 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 250 m). Click s = 250 m v i = 0v f = 10 m/s 20 kg f = 2 N A B F A = 12 N 60 o F H = 6 N How much work was done to overcome friction? k) 500 J What was the total work done? l) 1500 J Click Summarize the amounts of work done: n) 1) To accelerate the cart 1000 J 2) To raise the cart 0 3) To overcome friction 500 J 4) Total work done 1500 J (Cart was not raised) Remember that the force must be parallel to the distance in the Work Formula. (1500 J – 500 J) Click

7
Physics Ex-38 Question-2 A force of 110 N, acting 60 o from the horizontal, is applied to a 10 kg cart whose initial velocity is 4 m/s. The final velocity is 16 m/s. If the frictional force of 5 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 24 m). Click Was the cart at rest? a) No What was the applied force? b) 110 N [E 60 o N] What was the horizontal component of the applied force? c) 55 N right What was the frictional force? d) 5 N left Click What was the resultant force? e) 50 N right Click (While traveling from point-A to point-B) (F R = F A – f = 55 N – 5 N = 50 N) s = 24 m v i = 4 m/sv f = 16 m/s 10 kg f = 5 N A B F A = 110 N 60 o F H = 55 N “velocity not constant” Click

8
Physics Ex-38 Question-2 A force of 110 N, acting 60 o from the horizontal, is applied to a 10 kg cart whose initial velocity is 4 m/s. The final velocity is 16 m/s. If the frictional force of 5 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 24 m). s = 24 m v i = 4 m/sv f = 16 m/s 10 kg f = 5 N A B F A = 110 N 60 o F H = 55 N Click What was the acceleration of the cart? f) 5 m/s 2 What was the initial E K of the cart? g) 80 J What was the final E K of the cart? h) 1 280 J How much work was done on the cart? i) 1 200 J Click Work done on cart = total energy – energy lost to friction 1280 J – 80 J = 1200 J

9
Physics Ex-38 Question-2 A force of 110 N, acting 60 o from the horizontal, is applied to a 10 kg cart whose initial velocity is 4 m/s. The final velocity is 16 m/s. If the frictional force of 5 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 24 m). s = 24 m v i = 4 m/sv f = 16 m/s 10 kg f = 5 N A B F A = 110 N 60 o F H = 55 N Click How much work was done to overcome friction? j) 120 J What was the total work done? k) 1320 J Summarize the amounts of work done: l) 1) To accelerate the cart 1200 J 2) To raise the cart 0 3) To overcome friction 120 J 4) Total work done 1320 J (Cart was not raised) (1320 J – 120 J) Click

10
Physics Ex-38 Question-3 A 10 kg cart is traveling at 12 m/s towards the right. If a frictional force of 40 N stops the cart in a distance of 18 m, answer the following questions concerning the cart in going from point-A to point-B. Click Was the cart at rest? a) No What was the frictional force? b) 40 N left (or – 40 N) What was the resultant force? c) 40 N left (or - 40 N) What was the acceleration? d) - 4 m/s 2 Click (While traveling from point-A to point-B) s = 18 m v i = 12 m/s v f = 0 10 kg f = 40 N A B “velocity not constant” Click

11
Physics Ex-38 Question-3 A 10 kg cart is traveling at 12 m/s towards the right. If a frictional force of 40 N stops the cart in a distance of 18 m, answer the following questions concerning the cart in going from point-A to point-B. Click What was the initial E K of the cart? e) 720 J What was the final E K of the cart? f) 0 How much energy did the cart lose? g) 720 J What becomes of the energy lost by the cart? h) Click s = 18 m v i = 12 m/s v f = 0 10 kg f = 40 N A B Used to overcome friction (lost as heat and sound)

12
Physics Ex-38 Question-4 A hammer falls from a scaffold and 1.5 s later strikes the ground with a kinetic energy of 157.5 J. What is the weight of the hammer? Click

13
Physics Ex-38 Question-5 A projectile, whose mass is 800 g, is shot into the air with a velocity of 25 m/s, 42 o N of E. Determine the kinetic energy of the projectile one second after it is fired. Click

14
Physics Ex-38 Question-6 Starting from rest, a car reaches a velocity of 60 m/s in a distance of 120 m. Assuming the system is frictionless and knowing that the motor of the car produces a force of 3 x 10 4 N, calculate the mass of the car. Click

15
Physics Ex-38 Question-7 The mass of an electron is 1.67 x 10 27 kg. What work must be done on the electron in order to give it a speed of 2.5 x 10 7 m/s? Click

16
Physics Ex-38 Question-8 A bullet of mass 2 g, traveling at 500 m/s, is fired at a piece of wood. The bullet emerges from the wood with a speed of 100 m/s. If the retarding force of friction was 4800 N, calculate the thickness of the piece of wood. ? Click

17
Physics Ex-38 Question-9 Click What is the mass of a stone that is thrown in the air with a velocity of 1.1 m/s and with an initial kinetic energy of 0.0121 J? Click

18
Physics Ex-38 Question-10 Two vehicles, X and Y, are traveling at the same speed. Vehicle-X has twice the kinetic energy of vehicle-Y. What is the value of the following ratio? Click Mass of vehicle-X Mass of vehicle-Y

19
SSL Technologies.com/science

Similar presentations

Presentation is loading. Please wait....

OK

1. What is the weight of a 3.40 kg mass?

1. What is the weight of a 3.40 kg mass?

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on tunnel diode operation Ppt on garbage collection in java Ppt on women rights in islam Download ppt on transformation of energy Download ppt on coordinate geometry for class 9th Ppt on word association test psychology Ppt on tri gate transistor.pdf Ppt on youth power in india Ppt on history of badminton sport Ppt on layer 3 switching hub