A block of mass M slides along a horizontal table with speed v0. At x=...
The Problem
A block of mass M is sliding along a horizontal table with an initial speed v0. When it reaches x=0, it hits a spring with a constant k and begins to experience a friction force. The coefficient of friction is variable and is given by £=bx, where b is a constant. We need to find the loss in mechanical energy when the block first comes momentarily to rest.
Solution
To find the loss in mechanical energy, we need to calculate the work done by the friction force. The work done by a variable friction force can be found by integrating the product of the force and displacement over the entire path.
Calculating the Work Done
Let's consider a small displacement dx at a distance x from the starting point. The friction force acting on the block can be calculated as f = £mg = bmxg. The work done by this friction force over the small displacement dx is given by dW = f dx.
Since f = bmxg, we have dW = bmxg dx.
Integrating both sides of the equation, we get W = ∫(0 to x) bmxg dx.
Evaluating the integral, we have W = (1/2) bmgx^2.
Calculating the Potential Energy
When the block first comes momentarily to rest, its displacement x is maximum. At this point, all the initial kinetic energy is converted into potential energy stored in the spring.
The potential energy stored in a spring is given by PE = (1/2) kx^2.
Calculating the Loss in Mechanical Energy
The loss in mechanical energy is the difference between the initial kinetic energy and the potential energy stored in the spring.
Initial kinetic energy = (1/2) Mv0^2
Potential energy stored in the spring = (1/2) kx^2
Loss in mechanical energy = (1/2) Mv0^2 - (1/2) kx^2
Substituting the value of x from the equation x = v0^2 / (bg), we have
Loss in mechanical energy = (1/2) Mv0^2 - (1/2) kv0^4 / (b^2 g^2)
Simplifying further, we get
Loss in mechanical energy = (1/2) Mv0^2 (1 - v0^2 / (b^2 g^2))
Conclusion
The loss in mechanical energy when the block first comes momentarily to rest is given by (1/2) Mv0^2 (1 - v0^2 / (b^2 g^2)). This loss occurs due to the work done by the variable friction force.
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