Just want to double check my answers on this conceptual physics question :D?

Question

Ok so I have already worked these problems out but I'm just wondering what you guys get. Here are the problems:

Bill is in outer space on his way out of L.A. A force of 10 N pushes on Bill. Suddenly, when Bill has reached a speed of 250 mph, another 10 N force, directed opposite to the first force, also pushes on Bill. Now, with both forces acting on Bill, what happens to Bill's speed?

a) Bill will stop.
b) Bill will slow down.
c) Bill will speed up.
d) Bill will continue moving at a constant velocity.

Number 2:

Two blocks of masses 20 kg and 8 kg are connected together by a light string and rest on a frictionless level surface. Attached to the 8 kg mass is another light string which a person uses to pull both blocks horizontally. If the two-block system accelerates at 0.6 m/s^2, then what is the tension in the connecting string between the blocks?

a) 12 N
b) 6.0 N
c) 20 N
d) 4.0 N

Please provide why you feel the answer is right. Feel free to send me an IM. :)

Answer 1

1) When the second force is applied, the net force is zero.
Then the acceleration is zero. Then the last speed does not change any more. Answer 1D

2) F1 = the force between the two masses and F2 = pull force.
Look at the 20 kg: F = ma so F1 = 20 kg x 0.6 m/s2 = 12 N
Answer 2A

Answer 2

For the first question you need to calculate what is the net force.
The total force is 10+ (-10) = 0. What happens to a body that no force act on it. Review the first law of motion.

For the second:
The 20 kg mass is moving with an acceleration of 0.6 m/s2, being pulled by the string.
Review the second law of motion. What is the force if you have the mass and acceleration.

Best answer?

Answer 3

1. d Since equal and opposite forces are acting on Bill the net force is zero.

2. The tension in the connecting string is what causes the 20 kg block to accelerate at 0.6 m/s^2

Thus T = m*a = 20*0.6 = 12

Answer 4

number 1

d)
with a net force of "0" acting on bill, he will not accelerate or decelerate or change his velocity vector at all

number 2

a)

the 8 kg mass is irrelevant

the string between the two must accelerate the 20 kg mass

if you observe only the forces acting on the 20 kg mass you can see (F=ma) that it is 20kg * .6 m/sec^2 or 12 N


AND, I just read Dr. J's answer, and I vote for it as best

providing the incite without having to give the answer is often a better help technique

Answer 5

Number one is d.
Number two is c.

Answer 6

isn't this cheating? i've noticed a lot of people do this and i think its cheating.