Solve Physics 11 Homework: Work to Slide Block on Shelf

[roxxyroxx]

Homework Statement

A 1.0 kg block of wood is lifted a distance of 1.5 m from the floor and placed on a wooden shelf. A steady horizontal force is then applied to the block, causing it to slide sideways along the shelf a distance of 1.5 m at a uniform velocity. The kinetic coefficient of friction for the block when it is on the shelf is 0.23.
How much work is done sliding the block along the shelf?

Homework Equations

W = F x d

The Attempt at a Solution

Is there some formula I should be using?
I tried W = F x d but that doesn't give me the right answer of 3.4 J . Help?

 

[hage567]

What are you using for F?

 

[thomate1]

I would approach this problem by first identifying the relevant variables and equations. In this case, the relevant variables are the mass of the block (m = 1.0 kg), the distance it is lifted (d = 1.5 m), and the coefficient of friction (μ = 0.23). The equation that relates these variables to the work done is W = F x d, where W is the work done, F is the applied force, and d is the distance over which the force is applied.

To find the force, we can use Newton's second law, F = ma, where a is the acceleration of the block. Since the block is moving at a constant velocity, the acceleration is zero, and therefore the force must also be zero. This may seem counterintuitive, but it is because the force of friction is equal and opposite to the applied force, canceling it out and allowing the block to move at a constant velocity.

Now, we can use the equation for work to solve for the work done. W = F x d = (0 N) x 1.5 m = 0 J. This result makes sense, as the block is moving at a constant velocity and therefore no work is being done on it.

If the question is asking for the work done against friction, we can use the equation W = μmgd, where μ is the coefficient of friction, m is the mass of the block, g is the acceleration due to gravity, and d is the distance over which the force is applied. Plugging in the values, we get W = (0.23)(1.0 kg)(9.8 m/s^2)(1.5 m) = 3.4 J. This is the answer given in the problem, so it appears that the question may have been asking for the work done against friction rather than the total work done.

In summary, to solve this problem, we first identify the relevant variables and equations, and then use those equations to find the appropriate answer. It is important to carefully read and understand the question before attempting to solve it.