- #1
Lord Anoobis
- 131
- 22
μ
A simple diagram shows a block of mass M on a horizontal table with a light cord running from it to the right over a pulley at the edge of the table. At the other end of the cord hangs a block of mass m, which is a height h above the ground. The problem is to derive an expression for the speed of the block m just before it hits the floor if the coefficient of friction for the block on the table is μk.
Firstly, let me say that my second attempt at this problem led to the correct answer. However, this is of little consolation, because the reason behind the error in the first effort and its subsequent rectification is not clear.
I used the tension in the cord ##T = \frac{mMg(μ_k + 1)}{m + M}## to determine the work done by that force, the friction the already incorporated. This work, ##W_T## is a negative quantity, so
##W_T = -\frac{mMgh(μ_k + 1)}{m + M}##
Then followed ##K + W_T = U_g## leading to
##\frac{1}{2}mv^2 - \frac{mMgh(μ_k + 1)}{m + M} = mgh##
And some manipulation leads to the incorrect expression. Changing the negative sign to positive in the above equation gives the correct answer. What is the error here?
Homework Statement
A simple diagram shows a block of mass M on a horizontal table with a light cord running from it to the right over a pulley at the edge of the table. At the other end of the cord hangs a block of mass m, which is a height h above the ground. The problem is to derive an expression for the speed of the block m just before it hits the floor if the coefficient of friction for the block on the table is μk.
Homework Equations
The Attempt at a Solution
Firstly, let me say that my second attempt at this problem led to the correct answer. However, this is of little consolation, because the reason behind the error in the first effort and its subsequent rectification is not clear.
I used the tension in the cord ##T = \frac{mMg(μ_k + 1)}{m + M}## to determine the work done by that force, the friction the already incorporated. This work, ##W_T## is a negative quantity, so
##W_T = -\frac{mMgh(μ_k + 1)}{m + M}##
Then followed ##K + W_T = U_g## leading to
##\frac{1}{2}mv^2 - \frac{mMgh(μ_k + 1)}{m + M} = mgh##
And some manipulation leads to the incorrect expression. Changing the negative sign to positive in the above equation gives the correct answer. What is the error here?