Answer to Question #290654 in Optics for Mark

Question #290654

Activity Title: Mechanical Energy


Learning Target : To determine the mechanical energy of a system. Consider a simple pendulum with a bob mass 4.2 kg



Questions:


1. If the speed of the bob at the lowest point B is 0.84 m/s , What is its kinetic energy at this point?


2. If points A and C are the highest points reached by the bob, What is the potential energy if h= 0.036m at these points.?


3. What is the kinetic energy of the bob points A and C ?


4. What is the potential energy at point B.?


5. What is the total mechanical energy at point A.?


6. What is the total mechanical energy at point B.?

1
Expert's answer
2022-01-26T17:46:32-0500

Gives

M=4.2kg

Point B velocity

VB=0.84m/secV_B=0.84m/sec

Point A and C hiegh


h=0.036m

Part(a)

Kinetic energy of lowest point

KB=12mv2K_B=\frac{1}{2}mv^2


KB=0.5×4.2×(0.84)2=1.48176JK_B=0.5\times4.2\times(0.84)^2=1.48176J

Part(b )

Point (A) and Point (B) potential energy

PE=mghPE=mgh

PE=4.2×9.8×0.036=1.48176J4.2\times9.8\times0.036=1.48176J

PEA=PEC=1.48176JPE_{A}=PE_{C}=1.48176J

Part (c)

Kinetic energy of point ( A)and (c)

KEA=KEC=0KE_{A}=KE_{C}=0

At Point( a) and point( c) kinetic energy equal zero

part(d)

Potential energy at point B

PEB=0JPE_{B}=0J

At Point (B) potential energy is zero

part(e) total machanical energy point (A)

(Machanical energy )A_A =

KEA+PEBKE_A+PE_B

Machanical energy at point (A) =0+mgh

ME=mgh

ME=1.48176J

Part(f)

Machanical energy at point (B)

ME=KEB+PEBME=KE_B+PE_B

ME=12mv2+0ME=\frac{1}{2}mv^2+0

Mechanical energy at point B

MEB=1.48176+0=1.48176JME_B=1.48176+0=1.48176J


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Comments

Cleo Elizalde
26.05.22, 00:08

Thankyou for this,i understand alot now!

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