What Is the Initial Speed of a Block Given Friction and Distance?

[schyuler2]

Homework Statement

Find the initial speed of a block which travels 10m along a horizontal surface if $$\mu$$= 0.40 between the block and the surface before stopping.

Homework Equations

$$\Sigma$$W = 1/2mv_B² - 1/2mv_A² + mgy_B - mgy_A

$$\Sigma$$W = W * dAB * cos (W, dAB)

f = $$\mu$$* N
N= mgsin$$\theta$$

The Attempt at a Solution

so far i have:
$$\Sigma$$W = 0
$$\Sigma$$W = W_N + W_W + W_f

and

$$\Sigma$$W = W_W * dAB * cos (270)
$$\Sigma$$W = W_W * 10m * 0


not sure if I'm doing this right or where to go from here

 

[qnney]

This problem requires an equation that you haven't included yet.

So we'll start with:
$$\Sigma$$F = ma_x
$$\Sigma$$F = Px + fx + Nx + Wx
ma_x = Px + fx+ Nx + Wx

$$\mu$$ = -a / g
Try to solve it from here.

 

[kerimek]

.

I would approach this problem by first identifying the known quantities and variables. The known quantities in this problem are the distance traveled (10m), the coefficient of friction (\mu = 0.40), and the acceleration due to gravity (g = 9.8 m/s^2). The variable we are trying to find is the initial speed of the block.

Next, I would draw a free body diagram of the block to visualize the forces acting on it. The block experiences a normal force (N) and a frictional force (f) in the opposite direction of motion. Additionally, there is the force of gravity acting downwards. Using Newton's second law, we can write the equation \Sigma F = ma, where \Sigma F is the sum of all forces acting on the block.

In this case, we can write \Sigma F = ma as follows:

\Sigma F = N - f - mg = ma

We can also write the equation for friction as f = \mu N. Substituting this into the equation above, we get:

N - \mu N - mg = ma

Simplifying this equation, we get:

N = (m + \mu m)g

Now, we can use the equation for work-energy to solve for the initial speed of the block. We can write this equation as:

\Sigma W = 1/2mvB^2 - 1/2mvA^2 + mgyB - mgyA

Since the block starts and stops at the same height, we can eliminate the terms involving y from this equation. Also, since the block starts from rest, vA = 0. Therefore, our equation becomes:

\Sigma W = 1/2mvB^2 - mgyB

We can also write the work done by friction as Wf = f * d, where d is the distance traveled. Substituting this into the equation above, we get:

\Sigma W = 1/2mvB^2 - \mu N * d

Substituting our equation for N from earlier, we get:

\Sigma W = 1/2mvB^2 - \mu (m + \mu m)g * d

Since the block starts and stops at the same height, the change in gravitational potential energy is zero. Therefore, we can simplify this equation further to:

0 = 1/2mvB^2