Find the acceleration of each block

[nns91]

Homework Statement

1. A 2kg block sits on a 4kg block that is on a frictionless table. The coefficients of friction between the blocks are 0.3 (the static) and 0.2 (the kinetic).

a. what is the max F that can be applied to the 4kg block if the 2kg block is not to slide.
b. If F is half this value, find the acceleration of each block and the force of friction acting on each block.

2. m₁ =250g is at rest on a plane that makes an angle of 30 degree above the horizontal. The coefficient of kinetic friction between the block and the plane is 0.100. The block is attached to a second block of m₂= 200g that hangs freely by a string that passes over a frictionless and massless pulley. When the 2nd block has fallen 30cm, its speed is ?
(a) 83 cm/s
(b)48 cm/s
(c)160 cm/s
(d) 59 cm/s
(e) 72cm/s

Homework Equations

F=ma, Fx= coeff * F_n

The Attempt at a Solution

So I got part a right which is 17.7N. However, I got part b wrong.

1/2 of F= 8.85N.

F_s,2 = coeff * F_n = 0.3.2.9.81= 5.9 N Is this right ??

I got marked wrong on this one but cannot figure out why.
2. So I draw FBD and wrote Newton's 2nd law for each block and get

m₂ - F_n*coeff of friction -m₁g*sin$$\theta$$ =(m₁ +m₂)*a.

So I got a and use V_f² = 2a$$\Delta$$x

I got around 14...cm/s so I chose c but it's wrong. Can you guys help me ??

 

[alphysicist]

Homework Statement

1. A 2kg block sits on a 4kg block that is on a frictionless table. The coefficients of friction between the blocks are 0.3 (the static) and 0.2 (the kinetic).

a. what is the max F that can be applied to the 4kg block if the 2kg block is not to slide.
b. If F is half this value, find the acceleration of each block and the force of friction acting on each block.

2. m₁ =250g is at rest on a plane that makes an angle of 30 degree above the horizontal. The coefficient of kinetic friction between the block and the plane is 0.100. The block is attached to a second block of m₂= 200g that hangs freely by a string that passes over a frictionless and massless pulley. When the 2nd block has fallen 30cm, its speed is ?
(a) 83 cm/s
(b)48 cm/s
(c)160 cm/s
(d) 59 cm/s
(e) 72cm/s

Homework Equations

F=ma, Fx= coeff * F_n

The Attempt at a Solution

So I got part a right which is 17.7N. However, I got part b wrong.

1/2 of F= 8.85N.

F_s,2 = coeff * F_n = 0.3.2.9.81= 5.9 N Is this right ??

No; the formula

$$f_{\rm s,max} = \mu_s N$$
gives the maximum value of the static frictional force (right when the top object is about to start sliding along the bottom object). However, with this smaller force the object will not slide, and so the static frictional force will simply adjust itself to whatever it needs to be to keep the surfaces from sliding.

So you don't need a frictional force formula; just use the fact that both boxes have the same acceleration to draw a single force diagram. What does that give?

 

[nns91]

So frictional force will equal and opposite to the Force F ?

 

[alphysicist]

So frictional force will equal and opposite to the Force F ?

No; the frictional force will act on the boxes in such a way so that their accelerations are equal (by slowing the bottom box and moving the top box).

But here you do not need to worry about the frictional force to find the acceleration. The objects are moving together, so pretend they are one object. Draw a single force diagram for the combined object. What forces are acting on the combined object, and what does that give for the acceleration?

(After you find the acceleration, then find the frictional force by looking at the force diagram for just one of the blocks.)