Solving Work and Energy Problems

In summary, the conversation discusses three physics problems involving the concepts of kinetic energy, potential energy, and work. The first problem involves finding the kinetic energy of a golf ball at its highest point after being hit from a tee, and the second problem involves finding the speed of a cyclist as she crests a hill without pedaling. The third problem involves calculating the average power expended by a person while making homemade ice cream. Various equations and formulas are used to solve these problems, such as the equations for kinetic and potential energy, as well as the formula for power (work divided by time). However, some mistakes were made in the calculations, such as not including the mass in the equation for kinetic energy and not properly setting up the equation for the
  • #1
chipsdeluxe
10
0
1.A 50.3-g golf ball is driven from the tee with an initial speed of 40.6 m/s and rises to a height of 29.9 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 6.17 m below its highest point?

a. KE=1/2mv^2
i used this formula to find the KE but i keep getting the wrong anwer
KE=1/2(0.0503 kg)(40.6)^2=41.5 but this is the wrong answer. i;m not sure why it's wrong...
b. i used the formula:
KEinitial+PEinitial=KEfinal+PEfinal
PE initial=0 cause the initial height is 0 and KE final is 0 because the final velocity is 0. so the i get: KEinitial=PEfinal and for final height i get 11. but i don't think that i did this correctly though

2. A cyclist approaches the bottom of a gradual hill at a speed of 27.0 m/s. The hill is 7.29 m high, and the cyclist estimates that she is going fast enough to coast up and over it without pedaling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.

i used KEinitial+PEinitial=KEfinal+PEfinal and solved for the final velocity and i got 14.8 but that is wrong

3. A person is making homemade ice cream. She exerts a force of magnitude 26.4 N on the free end of the crank handle, and this end moves in a circular path of radius 0.271 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.16 s, what is the average power being expended?

i think you would use the equation P=Work/Time in this problem but other than that, I'm not sure where to start.

any help on any of the problems would be great. thanks so much
 
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  • #2
chipsdeluxe said:
1.A 50.3-g golf ball is driven from the tee with an initial speed of 40.6 m/s and rises to a height of 29.9 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 6.17 m below its highest point?
a. KE=1/2mv^2
i used this formula to find the KE but i keep getting the wrong anwer
KE=1/2(0.0503 kg)(40.6)^2=41.5 but this is the wrong answer. i;m not sure why it's wrong...
What you've calculated is just the initial KE of the ball. Use conservation of energy to find the KE at the highest point.
b. i used the formula:
KEinitial+PEinitial=KEfinal+PEfinal
PE initial=0 cause the initial height is 0 and KE final is 0 because the final velocity is 0. so the i get: KEinitial=PEfinal and for final height i get 11. but i don't think that i did this correctly though
Careful! The KE would be zero at the highest point only if the ball were hit straight up. (That's why they gave you its maximum height in order to solve part a. ) But you can use conservation of energy.
2. A cyclist approaches the bottom of a gradual hill at a speed of 27.0 m/s. The hill is 7.29 m high, and the cyclist estimates that she is going fast enough to coast up and over it without pedaling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.
i used KEinitial+PEinitial=KEfinal+PEfinal and solved for the final velocity and i got 14.8 but that is wrong
The idea is correct. Show the details of your work and we can find the problem.
3. A person is making homemade ice cream. She exerts a force of magnitude 26.4 N on the free end of the crank handle, and this end moves in a circular path of radius 0.271 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.16 s, what is the average power being expended?
i think you would use the equation P=Work/Time in this problem but other than that, I'm not sure where to start.
Hint: The work is force X distance. What's the work done to turn the handle once?
 
  • #3
1.A 50.3-g golf ball is driven from the tee with an initial speed of 40.6 m/s and rises to a height of 29.9 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 6.17 m below its highest point?

the KE at the highest point is...

KEi + PEi = KEf + PEf

PEf = mgh
PEi = 0
KEi= 1/2mv^2
KEf= 1/2mvf^2

replace thse formulas and solve for KEf...

and b is the same thing differnet heights... good luck
 
  • #4
thanks for the help guys but I'm still having a little troble
1. A 50.3-g golf ball is driven from the tee with an initial speed of 40.6 m/s and rises to a height of 29.9 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 6.17 m below its highest point?

i got the correct answer for part b but I'm having troble with part a
i used:
KEi+PEi=KEf+PEf
PEi=0 so it's KEi=KEf+PEf
1/2(40.6)^2=KEf+(9.8)(29.9)
KE=531.16
what did i do wrong?

2. A cyclist approaches the bottom of a gradual hill at a speed of 27.0 m/s. The hill is 7.29 m high, and the cyclist estimates that she is going fast enough to coast up and over it without pedaling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.

for this problem i did:
vf=square root(Vo^2+2g(Hinitial-Hfinal)
=square root(27^2+2(9.8)(7.29)) and i get 29.5 which is the wrong answer
 
  • #5
1/2(40.6)^2

? where is your mass?

1/2 m v^2


+(9.8)(29.9)

again , where is ur mass, u can't just cancel them out because you don't have a mass term on your KEf ... well you would if you break it down but you should not break it down... calculate it with the masses
 
  • #6
oh,i forgot the mass. i usually don't include the mass because they have have it and i just cancel it out. thanks for the help
 
  • #7
chipsdeluxe said:
2. A cyclist approaches the bottom of a gradual hill at a speed of 27.0 m/s. The hill is 7.29 m high, and the cyclist estimates that she is going fast enough to coast up and over it without pedaling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.
for this problem i did:
vf=square root(Vo^2+2g(Hinitial-Hfinal)
=square root(27^2+2(9.8)(7.29)) and i get 29.5 which is the wrong answer
Hinitial-Hfinal is negative.
 
  • #8
thanks again for the help
 

Related to Solving Work and Energy Problems

1. What is work in terms of physics?

In physics, work is defined as the force applied to an object multiplied by the distance over which the force is applied.

2. How is work related to energy?

Work and energy are closely related concepts in physics. Work is the measure of the transfer of energy from one object to another, or the change in energy of an object. Work can be done on an object to increase its energy, or an object can do work and decrease its energy.

3. What are some common units for work and energy?

The most common unit for work and energy is the joule (J), which is equal to one newton-meter (N*m). Other units include kilojoules (kJ) and ergs (erg). For energy, the calorie (cal) and kilocalorie (kcal) are also frequently used.

4. How is the work-energy theorem used in problem solving?

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This theorem can be applied in problem solving to find the work done on an object or to calculate the change in its kinetic energy.

5. Can you give an example of a work and energy problem?

Yes, an example of a work and energy problem could be: A 5 kg object is lifted 10 meters off the ground. If the force applied is 50 newtons, what is the work done on the object? Using the equation W = F*d, the work done would be 500 joules.

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