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Part B Rank each pendulum on the basis of the maximum kinetic energy it attains after release. Rank from largest to smallest To rank items as equivalent, overlap them. h 60 cm h 60 cm. h 45 cm h 30 cm 5 cm m 34kg m 2 kg m 1 m 3 kg m 8 kg largest Submit Hints My Answers Give Up Review Part Part C Rank each pendulum on the basis of its maximum speed Rank from largest to smallest To rank items as equivalent, overlap them. h 60 cm h 45 cm h 30 cm h 30 cm 15 cm m -4 kg largest Reset Help h 30 cm m 32 kg smallest Reset Help h 60 cm. m 1 kg smallest

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Answer #1
Concepts and reason

The concepts used to solve this problem are the conservation of mechanical energy in a simple pendulum.

Initially, use the concept of conservation of mechanical energy to find the potential energy and rank the potential energy that is equated to kinetic energy in the descending order.

Finally, use the expression for velocity of a simple pendulum and rank the velocities from the highest to lowest order.

Fundamentals

A simple pendulum is the one that can be considered to be a point mass suspended from a string or a rod of negligible mass.

In a simple pendulum with no friction, the total mechanical energy is conserved.

The total mechanical energy is the sum of kinetic energy and gravitational potential energy. Therefore, as the pendulum swings back and forth, there is a constant exchange between the kinetic and potential energies.

Therefore,

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Here, is the mass of the pendulum bob, is the acceleration due to gravity, is the length of the pendulum, and is the velocity of the pendulum.

(a)

According to the law of conservation of energy, the maximum kinetic energy is equal to the maximum potential energy.

Case:1

a)The potential energy is given as follows:

U, = mgh

Substitute for , 9.8m/s?
for , and for in the above expression.

U, = mgh
+(2ks)(9.8m/s)/ 60cm ( 19 cm
=11.76J

Case: 2

b)The potential energy is given as follows:

Ug = mgh

Substitute for , 9.8m/s?
for , and for in the above expression.

Un = mgh
= (1kg)(98m/=)co cm ( in time)
= 5.88J

Case: 2

c)The potential energy is given as follows:

Ug = mgh

Substitute for , 9.8m/s?
for , and for in the above expression.

U= mgh
= (4kg)(9.8m/s2)| 30cm( 10cm)
=11.76J

Case: 4

d)The potential energy is given as follows:

U. = mgh

Substitute for , 9.8m/s?
for , and for in the above expression.

Un = mgh
=(8kg)/(2.8m/s)( 15cm ( to crom
= 11.76J

Case: 5

a)The potential energy is given as follows:

Ug = mgh

Substitute for , 9.8m/s?
for , and for in the above expression.

Ug = mgh
=(3kg)(9.8m/s?)(45cm ( 10 cm)
= 13.23J

Case: 6

a)The potential energy is given as follows:

U = mgh

Substitute for , 9.8m/s?
for , and for in the above expression.

Un = mgh
= (2x3)(9.8m/s)(30cm (10.com)
1cm
)
= 5.88J

Therefore, arranging the potential energies in the descending order,

U, >U, TU, TU,>U, TU,

Since the maximum kinetic energy is equal to the maximum potential energy, the order of kinetic energy in the descending order can be written as given below:

K, >K, = K2 = K, >K, = KG

Here, denotes the kinetic energy.

(b)

The conservation of mechanical energy is given by the expression as follows:

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Re-arrange the above expression for velocity as follows:

v=J2gh

Case:1

a)The maximum speed is given as follows:

v=2gh

Substitute for and for in the above expression.

v; = /2gh
= 72(9.8m/sº)( 60cm ( lo com
= 3.42 m/s

Case: 2

b)The maximum speed is given as follows:

v2 = /2gh

Substitute for and for in the above expression.

V = 2gh
= 72(1.8m/s)( 45cm ( 10 cm)
= 2.96m/s

Case: 3

c)The maximum speed is given as follows:

V, = 2gh

Substitute for and for in the above expression.

V= 2gh
= /219.8m/s) 30cm (19cm)
= 2.42 m/s

Case: 4

d)The maximum speed is given as follows:

va = /2gh

Substitute for and for in the above expression.

va = V2gh
= /20.8m/5(30cm ( 19 cm)
= 2.42m/s

Case: 5

e)The maximum speed is given as follows:

Vs = 2gh

Substitute for and for in the above expression.

Vs = /2gh
= 72(.8m/=)/15cm ( locom
= 1.71m/s

Case: 6

f)The maximum speed is given as follows:

Vo = /2gh

Substitute for and for in the above expression.

Vo = 2gh
+72(.8m/s) socm( .com)
= 3.42 m/s

Therefore, arranging the maximum speed in the descending order,

<
=
<411
=

Ans: Part a

Thus, the maximum kinetic energy of the pendulum ranked in the descending order is K, >K, = K2 = K, >K, = KG
.

Part b

Thus, the maximum speed of the pendulum can be arranged in the descending order as v=V>v>V, = V, > vs
.

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