PHYS LB- Verification of Work-Energy Theorem PDF

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VERIFICATION OF WORK-ENERGY THEOREM
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EXPERIMENT IN VERIFICATION OF WORK-ENERGY THEOREM

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  PHYSICS 1433 GENERAL PHYSICS I LAB REPORT   NEW YORK CITY COLLEGE OF TECHNOLOGY 

 DEPARTMENT OF PHYSICS SPRING 2020

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VERIFICATION OF WORK-ENERGY THEOREM   

Touheda Khanom Prof. Lusik Hovhannisyan 04.29.2020

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Contents

1-0 OBJECTIVES

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1-1 THEORETICAL BACKGROUND

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1-2 PROCEDURE

6-7

1-3 DATA

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1-4 SAMPLE CALCULATIONS 1-5 QUESTIONS

7-10 10-12

1-6 CONCLUSION 

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OBJECTIVES The purpose of the laboratory was to determine the coefficient of the kinetic friction and verification of the Work-Energy Theorem. The aim is to understand the Work-Energy Theorem and perform measurements to evaluate the amount of work done on an object by a force of friction. To calculate potential and kinetic energies when an object slides on an inclined plane. THEORETICAL BACKGROUND The most important concept in this lab is the work-energy theorem. Energy is defined as the ability of a body to perform work. Energy can be in different forms such as mechanical, electrical, atomic, and nuclear energy. There are two kinds of mechanical energy. The energy that an object possesses by virtue of its position and interaction with another body is known as potential energy. The energy is conserved and the energy is never created nor destroyed. The law of conservative mechanical energy states that in an isolated system, the total mechanical energy of the system remains constant. According to Newton’s second law force is related to the rate of change of speed of the object. Therefore the work is done on the object by this net force as the object moves from initial position x1 to a final position x2 is given as W =

x2

∫F

dx (Formula for work done)

x1

From Newton’s second law, F = m dv dt

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W =

mv22 2

-

mv21 2

(Formula for work-energy theorem)

In the work kinetic energy theorem, it is convenient to subdivide the work into two categories: W con , the work is done by the conservative forces and W nonc the work done by the nonconservative forces. The kinetic energy theorem becomes W con + W nonc =

mv22 2

mv21 2

-

(Formula for work-energy theorem where conservative

and con-conservative forces included) W con = U 1 − U 2 (Formula for work of conservative forces between initial and final positions) 2

W nonc + U 1 − U 2 = mv22 W f = ( mgy 2 +

mv2 2 2

mv12 2

(Rewrite Work-Energy Theorem)

) − (mgy 1 +

mv12 2

) (Total Mechanical energy)

W f = − F k d ( Work done by kinetic friction) In the equation above, F k is the force of kinetic friction and the minus sign indicates that the force of friction is opposing the motion. The frictional force is F f = μk N (Kinetic Frictional force) W f = − μk mgd cosθ (Combining Normal and Coefficient Kinetic Friction) N = mgcosθ (Normal Force) μk =

mgsinθ N

=

mgsinθ mgcosθ

= tan θ (Coefficient of Kinetic Friction)

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The equation above can be used to determine the coefficient of kinetic friction μk by measuring angle θ0 at which the block slides down on the inclined plane at a constant speed. PROCEDURE 1. First, we measured and recorded the mass of the block and width of the flag. 2. We placed the block on the surface and measure and record the angle θ0 at which the block slides down on the inclined plane at a constant speed. 3. We repeated step 2 placing masses 100, 200, and 400 grams on top of the wooden block. Then recorded angle θ0 in each case. 4. We set up the angle θ of the adjustable inclined plane greater than θ0 and then set up the photogates. 5. We measured the distance between photogates along the inclined plane and the vertical distances between the base of the inclined plane and points of the inclined plane. 6. We connected the photogates to the computer and created an experiment in Data Studio Window. 7. We set up channel 1, selected blocked and then unblocked, and channel 2, selected blocked and then unblocked. 8. We record the time t1 and t1 where the glider is put in the starting point and measured times takes to pass the photogates between the photogates

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9. Then we repeated the steps 8 and 9 by adding masses of 100g and 200g on top of the wooden block. 10.We increased the angle θ of the inclined plane and repeat steps 8 through 10. DATA

SAMPLE CALCULATIONS In the experiment, I used the following formulas to get the values I put on the table. The formulas and calculations are shown below: GIVEN:

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Mass = .2192 kg Width of the flag = 0.03m

0 .03 Velocity (v 1 ) = 0.2020

Velocity = 0.149 m/s Velocity(v2) =

0 .03 0.0715

Velocity = 0.420 m/s

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E1 = E1 =

(0.2392)(0.149) 2 2

mv21 2

+ mgy 1

+ (0.2392kg)(9.8 m/ s2 ) (0.136 m) E 1 = .3215 J E2 =

E2 =

(0.2392)(0.420) 2 2

mv22 2

+ mgy 2

+ (0.2392kg)(9.8 m/ s2 ) (0.0894 m) E 2 = .2307 J

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ΔE = E 2 − E 1 ΔE = |.2307 − .3215 J | ΔE = .0856 J W f = − μk mgd cosθ W f = (.21 × .2392 × 9.8 × .180)Cos(15) W f = .0856 J % difference =

|0.0856− 0.0856| (0.0856+ 0.0856)/2

× 100%

% difference = 0 %

QUESTIONS

1. When the block moves an inclined plane, what kind of energy transformation takes place? - Kinetic energy is the energy transformation that takes place when the block moves an inclined plane. 2. Distinguish between static and kinetic friction. Explain why it is necessary for the block to move at a constant velocity to determine the coefficient of kinetic friction.

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Static friction is the applied force to prevent an object from moving. However, Kinetic friction has to overcome the maximum static friction to gain motion.

3. Step 8 of the procedure states that the block must be released from the same point for each trip. Why? -

The block must be released from the same point for each trip because if the velocity changes then the friction will also change.

4. Does your data confirm the expected relationship between the work done by the kinetic friction and the change in the mechanical energy of the block? State clearly what is expected and what your data indicates. - My data partially confirms the relationship between kinetic friction and the change in mechanical energy. The results for change in energy were positive while the changes in mechanical energy were negative.

5. A 2kg block slides down on an inclined plane and reaches the bottom with speed 4m/s. How much work does the force of friction do if the block starts from rest at a height of 1.5m? 2

E = mv2 + mgy = (4m/s)^2]/2+2kg*9.8m/s^2*1.5m  =45.4 J

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6. A ball of mass 0.2 kg is thrown straight upward with an initial speed of 15 m/s. When the ball returns to the same level, its speed is 13 m/s. How much work does air resistance do on the ball during its flight? mv21 2

W =

mv22 2

W =

(0.2kg )(13m/s) 2 2

-

-

(0.2kg )(15mls) 2 2

= -5.6 J 7. A car is moving horizontally with a speed of 20m/s when it breaks to a stop. Where does all the kinetic energy of the moving car go? - When the car with a speed of 20 m/s breaks to a stop the kinetic energy is transformed into heat.

CONCLUSION Based on our data and calculations, we discovered and understood the work-energy theorem. We also determined the coefficient of kinetic friction and verification of the work-energy theorem.

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