Two blocks on top of each other accelerating

1. Nov 5, 2004

FancyNut

I did this question several times and all I get for acceleration is negative 9.8...

static friction for the top block is .60 and kinetic friction for the bottom block is .20... The force is making both blocks move a distance of 5 meters starting from rest. They want to know What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

These are the equations I used....

acceleration is equal for both so $$a_t = a_b = a$$.

for the top block:

$$\sum F_x = F = m_t a + f_s$$

$$F = m_t a + u_s m_1 g$$

for the bottom block:

$$\sum F_x = F = m_b a + f_k$$

$$F = m_b a + u_k g (m_t + m_b)$$

I subtracted second equation from the first and get:

$$0 = a (m_t - m_b) + u_s m_1 g - u_k g (m_t + m_b)$$

$$a = - 9.8$$

then I used this equation:

$$x_f = x_i + v_i t + 1/2 a (t)^2)$$

$$5 = 1/2 (-9.8) t^2$$

$$t = 1.20408163$$

That time is of course wrong...

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Last edited: Nov 5, 2004
2. Nov 5, 2004

arildno

First of all:
You are using F as a force on the bottom block; but that is clearly false!
The driving force on the bottom block IS THE STATIC FRICTION FROM THE TOP BLOCK ON THE BOTTOM BLOCK!!

3. Nov 5, 2004

Staff: Mentor

Redo your equation for the bottom block. The applied force "F" acts only on the top block.

(arildno beat me again!)

4. Nov 5, 2004

FancyNut

Fixed it.

Thanks a lot. :)

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