Solving 5 Quick Homework Problems

  • Thread starter physicszman
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In summary: You have a car of mass "m" traveling at 20 km/hr. It has a certain amount of kinetic energy. If you want to slow it down to 10 km/hr, you have to take away half of that kinetic energy. If you want to slow it down to 5 km/hr, it will take away one third of its remaining kinetic energy. If you want to slow it down to 1 km/hr, it will take away half of its remaining kinetic energy.If you want to slow it down to zero, it will take away all of its remaining kinetic energy. That would mean that you would have to keep applying a force to it until it's velocity is
  • #1
physicszman
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1) A spring bathroom scale is designed to report the amount of upward force it's exerting on the objects touching its surface.

a) When you first step on the scale, you usually have some downward velocity because you "land" on the scale. As it slows your motion to bring you to rest, does teh scale report your correct weight, or does it report more than your weight or less than your weight.

*Greater than your weight.

b) If you stand motionless on one foot, rather than two feet, what fraction of your weight does the scale report?

*I believe it still reports your entire weight.

c) If you jump upward, what does the scake reoirt as you push yourself upward.

*Still your original weight?

d) You stand motionless on two identical scales, one foot on the left scale and one foot on the right. What can you say about the weights the two scales report.

*Their equal.

e) You stand motionless on two identical bathroom scales stacked ontop of each other. Each scale weights 10N. What weights do the scales report.(Bloomfield, 2001)

*The scale ontop of the scale reads your true weight and the one on the ground reads your weight + the one of the scale ontop of it.

2) Why would it be a bad idea to have your hand against the outfield wall while catching a batted ball? Explain (Beiser, 1992)

*It would be very painful because all of the force would be acting directly on your hand. There is no backward space to slow the ball down gradually.

3) The brakes of your car do work to slow the car down. If you slow from 20km/hr to 10km/hr, the brakes apply an average force over a certain distance. Assuming the brakes appy the same average force, how much further, as a friction of the first distance, will it take to slow the car from 100km/hr to 90km/hr? (Hint use the work energy principle and the definition of work)

*I don't have my book on me so I am just remembering this very rougly.

I know the work energy principle = W = (1/2mv^2) - (1/2mv2^2)

and work is defined to be the Force times distance of the cos of the angle.

I just don't know how to apply it to this problem since there is no mass.

4)At her highest point a 40kg girl on a swing is 2.0 m from the ground, while at her lowest point she is .08 m from the ground. What is her maximum speed? (Beiser, 1992)

*Help!

5)A 1000kg car strikes a tree at 30km/hr and comes to a stop in 15s. Find the intitial momentum and the average force on the car while it's being stopped ( Beiser, 1992)

*I think this problem requires the work enrgy principle as well.


Thanks for any help in advance.
 
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  • #2
In c) you are exerting force in addition to your weight in order to accelerate upward. The scale will report your weight plus that additional force.

In d) I would be inclined to say "they're equal" rather than "their equal" but I'll leave that to your English teacher. More precisely they each report half your weight.


3. The problem is not that there is no mass, but that there is no angle!

Kinetic energy is (1/2)m v2 where m is the mass and v the velocity. In slowing from 20 to 10 km/hr, your kinetic energy is reduced from (1/2)m(400) to (1/2)m(100)= 150m. Since you have not gone up or down, there is no change in potential energy, that change in energy must be accounted for by the friction of your tires on the road (and eventually becomes heat). Calling the Friction force F and the distance it takes to slow, that force is Fd. That is: Fd= 150m so
F= 150m/d. (I solved for F because the problem says " Assuming the brakes appy the same average force" so we need to use this in the second part.)

Slowing from 100 km/hr to 90 km/hr, kinetic energy reduces from
(1/2)m(10000) to (1/2)m(8100) so has reduced by (3439/2)m. Calling the distance required D, we must have FD= (3439/2)m. Since F= 150m/d from above, 150mD/d= (3439/2)m or D= (3439/300)d (the m's cancelled!). That's about 11.5 times as far!
 
  • #3
Oops, left out 4 and 5.

4, The 40 kg girl's highest point is 2 m off the ground so her potential energy (relative to the ground) is 2(40)g= 80g= 784 Joules. (g= 9.81 m/sec2 is the acceleration due to gravity). It is her highest point because she is no longer going upward but is not yet going downward: her speed is 0 and her kinetic energy is 0. Her total energy is 784 Joules.

When she is 0.08 off the ground, her potential energy (relative to the ground) is 0.08(40)g= 3.2g= 31.36 Joules. Since here total energy is conserved, the lost 784-31.36= 752.64 Joules must have gone into kinetic energy (ignoring friction loss, of course): (1/2)m v2= (1/2)(40)v2= 752.64 so v2= 37.6 and v= 6.1 m/s.

5. "A 1000kg car strikes a tree at 30km/hr and comes to a stop in 15s"
No, you don't actually need "energy-work" here. The change in momentum is enough. Momentum is, of course, mass*velocity so the car's momentum was 1000*30km/hr= 1000kg*30*100 m/3600 sec= 8.33 kgm/s. Since that momentum went down to 0 in 15 s, the average force was 8.33 kgm/s/15s= 55.6 kg m/s2= 55.6 Newtons.
 
  • #4
1. d) The correct answer is that the sum of the weights reported on the scales is your total weight. The two scales need not be showing the same weight.

Thought experiment: Imagine standing on one scale, and then stepping onto the other. Clearly, until your foot touches the second scale, all your weight is on the first one. You will then be transferring your weigh to the second scale until the first scale reads almost zero, just before your foot leaves it.
 
  • #5
For # 3 I don't understand how you got (3439/2).
 
  • #6
Originally posted by physicszman
For # 3 I don't understand how you got (3439/2).
I suspect Halls just made a mistake in copying a number:
Originally posted by HallsofIvy
Slowing from 100 km/hr to 90 km/hr, kinetic energy reduces from
(1/2)m(10000) to (1/2)m(8100) so has reduced by (3439/2)m. Calling the distance required D, we must have FD= (3439/2)m. Since F= 150m/d from above, 150mD/d= (3439/2)m or D= (3439/300)d (the m's cancelled!). That's about 11.5 times as far!
I'm sure he meant to write (1900/2) instead of (3439/2). That's about 6.3 times as far.
 
  • #7
Yea that's the answer I got too, thanks for the help!
 

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