Homework: Forces and Acceleration

[xasn23]

[SOLVED] Homework: Forces and Acceleration

1.)
While chopping down his father's cherry tree, Anthony discovered that if he swung the axe with a speed of 25 m/s, it would embed itself 2.3 cm into the tree before coming to a stop.
a.) If the axe head had a mass of 2.5 kg, how much force was the tree exerting on the axe head upon impact?

>>>I know how to get the force but I still have to take into account the 2.3 cm of embedding the axe made to the tree. The answer will come out negative right?

F=MA
F=(2.5 kg)(9.8m/s^2)
F=24.5Newtons

2.)
Jackson's favorite ride at Disneyland is the rollercoaster. The rollercoaster car and passengers have a combined mass of 1620 kg, and they descend the first hill at an angle of 45 degrees to the horizontal. With what force is the rollercoaster pulled down the hill?

>>> Again, I know how to find the force here, but the 45 degrees messes me up. Oh, but quick question, is the acceleration different here because what if the rollercoaster cable lines is a factor in the acceleration of the rollercoaster? Or is it still 9.8m/s^2??

 

[hage567]

For the first question, why would you say a = 9.8 m/s^2? Gravity is not involved in this question. You can use the 2.3 cm to find the deceleration of the axe as it hits the tree. You can do this because you also know the initial velocity and the final velocity. Look at your kinematic equations. Once you find "a" using that information, you can put it into F=ma.

 

[xasn23]

I'm not sure what the final and initial velocity is ...

I'm thinking that the final is 0 since the axe stops in the end...
But I'm still skeptical about the initial... how do I find the initial?

Please help? =)

 

[hage567]

Yes, the final velocity will be zero because the axe comes to a stop. The initial velocity is 25 m/s, as stated in the problem.

 

[xasn23]

ahh ok I see thank you! >:D<

 

[hage567]

Have you figured out the second question?

 

[xasn23]

... dangit ...

Wait ok so since I'm dealing with a 45 degree drop... does that mean I'll take the F=MA and divide it by 2?

I'm pretty sure that that's what I'm supposed to do...

 

[hage567]

Since there is the 45 degree incline, the acceleration of the car will not be 9.8 m/s^2. You need to find the component of gravitational force that causes the car to accelerate down the hill. Do you know about vector components? Have you drawn a diagram and labelled the forces present?

 

[xasn23]

no... I don't ...

I just learned this stuff a few hours ago.

 

[hage567]

Take a look at this site. See if the diagrams help. I'm sure there's some stuff in your textbook as well.

http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html#c3

 

[xasn23]

my textbook doesn't help me because it uses different situations... but it doesn't help me understand the 45 degrees part... do I find the acceleration first like so:

(sin45*1620)

or something of that sort... I'm still not sure how to.

I'd really appreciate it if you helped me out... I'm not just here for a freebie homework answer... it'd be awesome if I can understand this stuff.

 

[xasn23]

oh yeah.

I got both of them already.
The first one I have to use
Acceleration= [(Final Velocitysquared - Initial Velocitysquared)]/ 2*distance in m

so I changed 2.3 cm > .023m

And then
0-25^2 625m/s
A= _______ = _____ = 13,586.957 m/s
2 (.023m) .046m

For the second one. All I had to find out was the Component of gravitational force and the normal force. and what I did was

Normal Force= (1620kg)(9.8 m/s)/cos45

and then I found the Component of Gravitational force

C of GF = (sin45)(9.8m/s)(1620kg)

^__^ yeah I looked over the website you gave me again too and it finalized my answer. THANK YOU! >:D<

 

[xasn23]

Take a look at this site. See if the diagrams help. I'm sure there's some stuff in your textbook as well.

http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html#c3

I got both of them already.
The first one I have to use
Acceleration= [(Final Velocitysquared - Initial Velocitysquared)]/ 2*distance in m

so I changed 2.3 cm > .023m

And then
0-25^2 625m/s
A= _______ = _____ = 13,586.957 m/s
2 (.023m) .046m

For the second one. All I had to find out was the Component of gravitational force and the normal force. and what I did was

Normal Force= (1620kg)(9.8 m/s)/cos45

and then I found the Component of Gravitational force

C of GF = (sin45)(9.8m/s)(1620kg)

^__^ yeah I looked over the website you gave me again too and it finalized my answer. THANK YOU! >:D<