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Two objects are connected by a string of negligible mass

  • #1
Two objects are connected by a string of negligible mass.
The 10 kg block is placed on a
smooth table top 2.2 m above the floor, and
the 8 kg block hangs over the edge of the ta-
ble. The 8 kg block is then released from rest
at a distance of 1.1 m above the floor at time
t = 0.


A. Determine the acceleration of the 8 kg block
as it descends. The acceleration of gravity is
9.8 m/s2 .
Answer in units of m/s2.


B. How long dies it take the 8 kg block to strike
the floor?
Answer in units of s.


C. Describe the motion of the 10 kg block from
the time t = 0 to the time when the 8 kg block
strikes the floor.
1. Itmoves with an increasing acceleration.
2. It moves with a constant speed.
3. It moves with a decreasing acceleration.
4. It moves with a constant acceleration.
5. It stays still.


D. Describe the motion of the 10 kg block from
the time the 8 kg block strikes the floor to the
time the 10 kg block leaves the table.
1. Itmoves with an increasing acceleration.
2. It stays still.
3. It moves with a constant speed.
4. It moves with a decreasing acceleration.
5. It moves with a constant acceleration.


E. Determine the distance between the landing
points of the two blocks.
Answer in units of m.

I've tried this question multiple times and got it wrong.... :( I need help!
 

Answers and Replies

  • #2
kuruman
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Hi bahamut1111 and welcome to PF. Can you show us what answer(s) you gave and why you thought these were the correct answers? Then we may be able to point out where you went wrong and why. Start with question A.
 
  • #3
ok i got the first four parts but the fifth im still getting wrong......


A. a = F/m = 8*9.8 / (10+8) = 4.36 m/s^2

B. 1.1 = (1/2) 4.36 t^2

t = sqrt(2.2/4.36) = 0.711 s

C. 4. It moves with a constant acceleration.

D. 3. It moves with a constant speed.

E. Since the 8 kg block lands on the floor even with the edge of the table.
10 kg block travels at a horizontal speed of v = a t = 4.36 * 0.711 = 3.1 m/s
The 10 kg block lands v t = 3.1*0.711 = 2.2 m from the edge of the table so they land 2.2 m apart. But when i enter that it marks it wrong.....
 
  • #4
am i doing something wrong?
 
  • #5
kuruman
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Parts A-D are correct. You have correctly found that the speed of the 10 kg block when it leaves the table is 3.1 m/s. What you are doing wrong is that to find how far from the table it lands you multiply incorrectly by 0.711 s. The 0.711 s is the time the 10 kg block changes its speed from zero to 3.1 m/s while sliding on the table. You need to multiply the 3.1 m/s by the time the block is in flight, i.e. the time interval from when the block leaves the table until it hits the floor. Can you find the time of flight?
 
  • #6
im already blanking out....
 
  • #7
kuruman
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im already blanking out....
That's not very helpful. If you did parts A-D yourself, you should have a picture in your head of what is going on. Can you describe with just words how the 10 kg block moves from t = 0 until it hits the floor?
 

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