# Interacting systems problems

1. Oct 11, 2004

### quick

im trying to do these homework problems but really don't know where to begin. they deal with multiple systems and they seem really easy i just don't know the formula to apply. any help with these would be appreciated.
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1) A 1000 kg car pushes a 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N. What is the magnitude of the force of the car on the truck?
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2) The 1.0 kg block in the figure is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is 0.440 (static friction). What is the tension in the rope holding the 1.0 kg block to the wall? What is the acceleration of the 2.0 kg block?
http://s93755476.onlinehome.us/stuff/knight.Figure.08.27.jpg [Broken]
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3) The 100 kg block in figure takes 5.50 to reach the floor after being released from rest.What is the mass of the block on the left?
http://s93755476.onlinehome.us/stuff/knight.Figure.08.33.jpg [Broken]
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4) A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 30 m/s when a 60 kg skydiver drops out by releasing his grip on the glider. What is the glider's speed just after the skydiver lets go?

Last edited by a moderator: May 1, 2017
2. Oct 11, 2004

### Gopi Prashanth

As for your first question, we need more information but assuming that either of them are not moving still....

Consider the force diagram of your car...

<------O ------>Truck
4500 N

Since the car is going to be in equilibrium the forces should balance each other... hence the force applied by the car on the truck would be........?

As for the other questions, go ahead and draw force diagrams... it really helps in understandint the problem.. once you have the force diagrams.. the only formula you will ever need is Net Force = Mass * Acceleration :)

For the last question, it is slighly tricky.. conservation of momentum is the key...

Hint: As soon as the sky diver drops out he still has the same velocity as the glider... use Mo*Vo = M1*V1 + M2*V2.. if you think a little bit you dont need the formula for the answer...

3. Oct 11, 2004

### quick

ok ive figured out all of the problems except for the third one. i have set up a force diagram with gravity acting on both masses as well as both of the blocks acting on each other. i am unsure of the equation for the net force on the right block in terms of the left block. i was thinking m_1*g - m_2*g = m*a where m_2 is the right block of 100kg. then i tried to find the acceleration by using the equation for the position of an object at time t given constant acceleration. with intial x as 1 m, intial velocity at 0 and t is 5.5 s. i got -.066 m/s^2 but don't know if this is correct. can someone lead me to the right direction?

4. Oct 11, 2004

### Pyrrhus

The pulley is masless and frictionless right? if so then its the same tension for both blocks.

5. Nov 13, 2004