What is coefficient for static friction

Question

A 6-g bullet is shot into a 2.1-kg block of wood, where it becomes embedded. The block is initially at rest and the bullet has an initial velocity of 400 m/s. The block then travels 1.2 m horizontally. What is the coefficient of kinetic friction between the block and the surface?  

Answer 1

Conservation Of Momentum gives the momentum of the block and bullet combined at, say, a hundredth of a second after the bullet strikes the block as p = mass_bullet × v_0bullet.

From this, the Kinetic Energy of the two objects joined is K = p² ÷ [2(m_block+m_bullet)] or

m_bullet²v_0bullet² ÷ [2(m_block+m_bullet)].

The Friction Force does Work W_f = -fs equal to -µ_k(m_block+m_bullet)gs in stopping the slide of the block.

It then follows that ΔK=W_f or 0−m_bullet²v_0bullet² ÷ [2(m_block+m_bullet)] equals -µ_k(m_block+m_bullet)gs.

Isolate µ_k as m_bullet²v_0bullet² ÷ 2sg(m_block+m_bullet)² or

(0.006)²(400)²[kg²×m²/s²] ÷ 2(1.2)(9.8)(2.106)² [kg²×m²/s²] or 0.05521644377 equivalent to 0.055.

Answer 2

Let's first build the solution logically in reverse!

Steps:

1. To find coefficient of kinetic friction using the formula F= µ * N , we have to find how much frictional force was applied to stop the block with bullet.
2. To find the force F, we have to find the work done by the force. WHich is equal to the change in Kinetic Energy of the bullet.
3. The bullet embedded in the block, came to a complete stop. So, the change in kinetic energy is equal to the initial kinetic energy. We need to get the initial velocity of the block with the bullet.
4. To get the velocity of the block with bullet, we can use conservation of momentum with the bullet and block collision.

Now let's start from the last step and reach to our goal of finding the coefficient of Kinetic friction

First we have to consider the momentum conservation

mass of bullet, m1=6g =0.006 Kg initial velocity of the bullet, v1 =400 m/s

mass of block, m2 = 2.1 Kg initial velocity of the block, v2=0 m/s

final mass, m3= 2.106 Kg what is the final velocity of the blockwith bullet, v3?

Using Momentum conservation, m1* v1 + m2*v2 = m3*v3

# 0.006Kg*400m/s + 2.1*0Kg m/s = 2.106Kg*v3

# 2.4 Kg m/s =2.106Kg *v3

# v3 = 1.1396 m/s

We can calculate the kinetic energy of the block with bullet from this

Kinetic Energy is, E = 1/2 m v² = 0.5*m3 * (v3)^2

= 1.3675 Kg m²s^-2

= 1.3675 J

Let's say that the Frictional force that stopped the bullet is F,

then Work done by this force= Kinetic Energy of the Block with Bullet

Since the force is a frictional force, we cas say it was a constant force and work done by it is

work, W = F*x , where x is the distance travelled by the Block with bullet.

So, x= 1.2m

Now, F*x = W = Kinetic Energy = 1.3675 J

so, F* 1.2m =1.3675 J

so, F = 1.3675 J / 1.2 m

= 1.1396 Kg m s^-2

= 1.1396 N

We know the frictional force, F = µ * N

where µ is the coefficient of friction and N is the Normal Force

Since the surface is horizontal, the weight of the block with bullet is equal to the magnitude of the normal force. N = m3 * gravitational acceleration

So, F = µ* 2.106*9.8 N

so, µ = 1.1396/ 20.6388

= 0.055

Coefficient of kinetic friction is 0.055