1. ## Physics

I have a test soon and I don't understand these types of problems very well...can anyone please explain these?

1) A carton of eggs rests on the seat of a car moving at 22.5 m/s. What is the least distance in which the cart can be uniformly slowed to a stop if the eggs are not to slide? The coefficient of friction between the seat and the carton is 0.24

2) Two blocks are connected by string of negligible mass that passes over a frictionless pulley. The mass of the block on the table is 43.1 kg and the mass of the block in the air is 18.9 kg. Assume the table is frictionless. Find the acceleration of the two blocks and the tension in the string

Answers are 2.98 m/s^2 & 129N

3) 2 masses 5 kg and 10 kg are tied to the opposite ends of a massless rope, and the rope is hung over a massless and friictionless pulley. Find the acceleration of the masses

4) Block 1, 8 kg, is moving on a 33 degree incline ( coefficient of kinetic friction is 0.25. This block is connected to block 2, 22 kg, by a cord that passes over a massless and frictionless pulley. Find the acceleration of each block and the tension in the cord

Answers are 5.2 m/s^2 and 101N

2. Originally Posted by realintegerz
I have a test soon and I don't understand these types of problems very well...can anyone please explain these?

1) A carton of eggs rests on the seat of a car moving at 22.5 m/s. What is the least distance in which the cart can be uniformly slowed to a stop if the eggs are not to slide? The coefficient of friction between the seat and the carton is 0.24

[snip]
$\displaystyle F_{\text{net}} = ma$.

If carton of eggs are on the point of sliding then $\displaystyle F_{\text{net}} = - \mu N = - (0.24) (m) (9.8)$.

Therefore $\displaystyle m a = - \mu N = - (0.24) (m) (9.8) \Rightarrow a = - (0.24)(9.8) = - 2.352$.

a = - 2.352 m/s^2
u = 22.5 m/s
v = 0 m/s
x = ?

Solve for x using an appropriate motion-under-constant-acceleration formula.

3. Originally Posted by realintegerz
[snip]
2) Two blocks are connected by string of negligible mass that passes over a frictionless pulley. The mass of the block on the table is 43.1 kg and the mass of the block in the air is 18.9 kg. Assume the table is frictionless. Find the acceleration of the two blocks and the tension in the string

Answers are 2.98 m/s^2 & 129N

[snip]
Did you draw a force diagram? Equations of motion:

43.1 kg mass:

(43.1)(a) = T .... (1)

18.9 kg mass:

(18.9) (a) = (18.9)(9.8) - T .... (2)

Solve equations (1) and (2) simultaneously for a and T.

4. Originally Posted by realintegerz
[snip]
3) 2 masses 5 kg and 10 kg are tied to the opposite ends of a massless rope, and the rope is hung over a massless and friictionless pulley. Find the acceleration of the masses

[snip]
Did you draw a force diagram? Equations of motion:

10 kg mass:

10 a = (10)(9.8) - T .... (1)

5 kg mass:

5a = T - (5)(9.8) .... (2)

Solve equations (1) and (2) simultaneously for a.

5. Originally Posted by realintegerz
[snip]
4) Block 1, 8 kg, is moving on a 33 degree incline ( coefficient of kinetic friction is 0.25. This block is connected to block 2, 22 kg, by a cord that passes over a massless and frictionless pulley. Find the acceleration of each block and the tension in the cord

Answers are 5.2 m/s^2 and 101N
Did you draw a force diagram? Resolve the forces on the 8 kg mass? Equations of motion:

22 kg mass:

22 a = (22)(9.8) - T .... (1)

8 kg mass:

Perpendicular to slope: 0 = R - (8)(9.8) cos 33 => R = (8)(9.8) cos 33 .... (2)

Parallel to slope: 8a = T - (8)(9.8) sin 33 - (0.25) R .... (3)

Substitute equation (2) into equation (3) and solve equations (1) and (3) simultaneously for a and T.

6. I don't know what formulas to use to solve for a & t....

7. Originally Posted by realintegerz
I don't know what formulas to use to solve for a & t....
I'm not solving the equations for you as well.

If you're studying dynamics you must surely have been taught basic things like how to solve simultaneous equations.