Question

#4. A mass of 1.00 kg is sitting 10 m in the air above a spring of spring constant k=50 N/m. The mass is released and lands on the spring compressing it by a distance indicated by yf

a) What is the TE of this system?

b) What is the KE of the mass just as it makes contact with the spring?

c) What is the position of maximum compress (yf)?

#5. A 3.0 kg mass is sitting on a table attached by a string over a pulley to a 30 kg mass hanging off the table. The coefficient of friction between the mass on the table and the table is (mk=0.2). The mass hanging off the table falls 1m.

a) What is the change in PE for each mass?

b) How much work is done by friction?

c) What is the final KE for the entire system and what is the final velocity?

Answer 1

4Given mass,m=1kg

height ,h =10m

a)Total Energy =Potential Energy +Kinetic Energy

When object is at height 10m just about to drop ,there is no kinetic energy

so TE = PE =mgh

TE = 1*9.81*10

ANSWER: Total Energy = 98.1 J

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(b)When the body is about to make contact with the spring (h becomes zero)its Potential energy is completely converted into Kinetic energy

so KE = 98.1 J

ANSWER: KE = 98.1 J

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(c)When the spring is compressed and body comes to a stop,its KE is converted into potential energy of spring

so KE =PE of Spring

98.1 =1/2ky_f²

Given k =50N/m

y_f² = 98.1*2/K

y_f = 1.981 m

ANSWER: y_f = 1.981 m

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5)

Given m₁ = 3kg ,m₂ = 30kg

= 0.2

Falling distance =sliding distance,S =1m

a)change in Potential energy for each mass

mass m₁ does not have any change in potential energy ,only kinetic energy

so for m₁ , = 0J

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for mass m₂

= mgS

= 294.3 J

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b) Work done by friction

so the Potential energy from 5.a) is converted into KE of m₁ ,KE of m₂ and some energy is lost due to friction

so Workdone by friction = -Fr *S

where Fr is the frictional force =

W_f = -  

ANSWER:

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(c)Final KE for entire system

ANSWER:

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Final Velocity

ANSWER:

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