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.?

Expert's answer

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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