What is the Average Frictional Force Acting on a Sliding Bear?

[kappcity06]

i got most of this problem but i don't know what average frictional force means.

A 25 kg bear slides, from rest, 13 m down a lodgepole pine tree, moving with a speed of 6.4 m/s just before hitting the ground.
(a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide?-3185J
b)What is the kinetic energy of the bear just before hitting the ground? 512J
c)What is the average frictional force that acts on the bear?

 

[Norman]

i got most of this problem but i don't know what average frictional force means.

A 25 kg bear slides, from rest, 13 m down a lodgepole pine tree, moving with a speed of 6.4 m/s just before hitting the ground.
(a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide?-3185J
b)What is the kinetic energy of the bear just before hitting the ground? 512J
c)What is the average frictional force that acts on the bear?

Can you find the amount of work done by friction?

 

[kappcity06]

would that just be potential energy minus kentic energy

 

[Norman]

would that just be potential energy minus kentic energy

In general: Nope.

If there are only conservative forces working, energy would be conserved. That is to say that the energy before an event is the same as the energy after the event. Is friction a conservative force?

 

[HallsofIvy]

Well, in this problem, the kinetic energy at the top is 0 so the bear's "total energy" is equal to it's potential energy at the top of the tree. At the bottom of the tree the potential energy is 0 so the total energy is just kinetic energy plus the energy lost through friction.

I'm not sure what Norman meant by "In general" but in this case, yes, the energy lost to friction is potential energy (at the top of the tree) minus the kinetic energy (at the bottom of the tree). Of course, that is also equal to the distance the bear slid times the average friction force.

 

[sdekivit]

again use the energy balance that is valid here:

[tex] E_{gravitation} = E_{kinetic} + E_{friction}[/tex]

=

[tex] mgh = \frac{1} {2} mv^{2} + E_{friction}[/tex]

if you can set up the energy balance like this, you are well ahead in these kinds of problems. It's very important to realize what energy is at the beginning of the process and how this energy is converted during a process. In this case, we start with gravitational energy and end up with kinetic energy and heat lost to the surroundings due to friction. Then it's simply setting up an equation like this :)

After knowing this energy, you know the work done by the frictional force and you can calculate the frictional force use the formula W = F * s * cos 180 degrees (and thus frictional work is negative since the force is 180 degrees against the movement)

 

[kappcity06]

again use the energy balance that is valid here:

[tex] E_{gravitation} = E_{kinetic} + E_{friction}[/tex]

=

[tex] mgh = \frac{1} {2} mv^{2} + E_{friction}[/tex]

if you can set up the energy balance like this, you are well ahead in these kinds of problems. It's very important to realize what energy is at the beginning of the process and how this energy is converted during a process. In this case, we start with gravitational energy and end up with kinetic energy and heat lost to the surroundings due to friction. Then it's simply setting up an equation like this :)

After knowing this energy, you know the work done by the frictional force and you can calculate the frictional force use the formula W = F * s * cos 180 degrees (and thus frictional work is negative since the force is 180 degrees against the movement)

this did not work i woulnd up with enegry of fricion=2673. then i but it into the equation w=fscos180 and got -34749?

 

[Office_Shredder]

The energy of friction is the total work done by friction, so you would plug that in for w. Then you would want to solve for force

 

[kappcity06]

alright thanks

 

[kappcity06]

posirtve or neagitive i didnt know friction could be negative

 

[Office_Shredder]

When you're talking about work, it can either be positive or negative. Essentially, if you're doing work to make something move, it's positive. If you're doing work to prevent something from moving, it's negative. So for w, you would technically be plugging in the negative value of energy of friction. However, the cos(180) is also negative, so they cancel.

Your answer should be positive

 

[kappcity06]

yea u were right. thanks

 

[sdekivit]

this did not work i woulnd up with enegry of fricion=2673. then i but it into the equation w=fscos180 and got -34749?

if you do the calculation correctly, this approach always works ;)

You should bear in mind that energy has the unit J and work too.