Questions & Answers on Force of Impact and Friction

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The discussion focuses on physics questions related to force, impact, and friction. For the first question, the acceleration of the person on roller skates pushing against a wall is determined to be zero, as the wall does not move. In the second question, the force acting on the hand during a karate chop is confirmed to be 4000 N, matching the force delivered to the board. The third question concludes that during a collision, the tiny car experiences the largest force of impact, despite both vehicles experiencing equal forces. Lastly, the frictional force acting on a bear sliding down a tree at constant velocity is identified as 3000 N, based on the equilibrium of forces.
MCS
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Hi,
I have a few more questions. Thanks for helping :-D

1) A person on roller skates stands next to a wall and pushes the wall with a force of 25 N. If the person has a mass of 50 kg what is the acceleration?
a.zero
b.0.5 m/s/s
c.2 m/s/s
d.4 m/s/s
I think it is A...

2) A karate chop delivers a blow of 4000 N to a board that breaks. The force that acts on the hand during this event is actually______ N.
a.0
b.2000
c.4000
d.8000
I think the answer C.

3)A very large truck and a tiny car traveling at the same speed collide head on. Which of the following is true?
a. the large truck will experience the largest force of impact.
b. the tiny car will experience the largest force of impact.
c. the force of both truck and car will be the same.
d. The truck will accelerate more than the tiny car as a result of the collsion force.
I think the car would accelerate more...The answer is B...

4)A 300 kg bear grasping a vertical tree slides down at a constant velocity. The friction between the tree and bear is ____N.
a.30
b.300
c.3000
d.greater than 3000

I have no idea on this one :-/ how would I go about figuring it out?

Thanks
 
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for 4, its a matter of considering the eqn:
sum of forces F=ma= mass*acceleration(due to gravity)-frictional force. since the bear is not accelerating, the sum then must be zero which means
frictional force=m(a).

Use the same formula f=m*a to look at problem 1 again.
 
So # 4 would be C?
 
MCS said:
So # 4 would be C?

Yes, it would.
 
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