Solution to Force & Friction: Find Time for 4kg Block to Drop 7m

[liquidheineken]

1. A 5 kg block rests on a horizontal table, attached to a 4 kg block by a light string as shown in the figure. The acceleration of gravity is 9.81 m/s 2 .

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The coefficient of kinetic friction between the block and the table is 0.3, find the time it takes for the 4 kg mass to fall 7 m to the floor if the system starts from rest. Answer in units of s.

2. Homework Equations
5 kg block = B
4 kg block = A
u = 0.3
g=9.81 m/s^2
Ff = Force of Friction = u*mass*acceleration
F = Force = mass*acceleration

The Attempt at a Solution

I started by finding the acceleration of the 5kg block using the following equation:
FB = mB * a = FA - Ff

Solving for FA:
FA = (4 kg)(9.81 m/s^2) = 39.24 N

solving for Ff:
Ff = (0.3)(5 kg)(9.81 m/s^2) = 14.715 N

solving for acceleration (a):
(5 kg) * a = (39.24 N) - (14.175 N)
(5 kg) * a = 25.065 N
a = (26.065 N) / (5 kg) = 4.905 m/s^2

solving for time (t) to reach 7 meters:
(a*t^2)/2 = 7 m
a*t^2 = 14 m
t^2 = (14 m) / (a)
t = sqrt [(14 m) / (a)] ≈ 1.689 seconds

The answer comes back as incorrect when I submit it. Any help would be appreciated.

 

[Merlin3189]

I think you have not taken account of the fact that the blocks are joined by the string.

You assume the force on block B = the gravitational force on block A - the frictional force on block B :this is not correct.

You mark on the diagram a letter T , which I assume is the tension in the string, but you do not use this. (Personally, I would also put on the diagram any forces I do use, such as F_A , F_B and F_f)

Remember both blocks move and you need to take account of the acceleration of block A.

 

[liquidheineken]

ok I solved the problem. My original equation was wrong. Apparently since both masses are moving with the same acceleration in this system, and there are only so many Newtons to go around, the equation used to find the acceleration looks more like the following:

Fsystem = (mB+mA) * a = FA - Ff

 

[liquidheineken]

I think you have not taken account of the fact that the blocks are joined by the string.

You assume the force on block B = the gravitational force on block A - the frictional force on block B :this is not correct.

You mark on the diagram a letter T , which I assume is the tension in the string, but you do not use this. (Personally, I would also put on the diagram any forces I do use, such as F_A , F_B and F_f)

Remember both blocks move and you need to take account of the acceleration of block A.

I just copied and pasted this problem out of my homework. That T was already marked. Is there an equation which can relate tension to the solution here?

 

[Merlin3189]

Fine. I also solved it as a combined mass, but mentioned the T because you chose to solve for each body separately (which is also ok.) Then T helps because the net force on A is M_A.g - T and on B is T - F_f giving two equations and when you add them to eliminate T, the result is the equation you just gave.