Kinematics - spring + friction

[nahanksh]

Homework Statement

http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam2/fa09/fig8.gif
A small block having a mass of 2 kg is in contact with an ideal spring of relaxed length 1 m and spring constant k = 100 N/m . The spring is compressed to a length of 0.5 m. The block is released from rest at x = 0.5 m. At x = 1 m the mass leaves the spring and comes to rest at x = 2 m. Throughout its entire motion the block slides on a rough surface with a coefficient of kinetic friction μk .

Q1) The maximum acceleration of the block occurs the instant the block begins to move.
True/False


Q2) What is the coefficient of kinetic friction of the surface?

Homework Equations

The Attempt at a Solution

For the first question, i thought the acceleration is at maximum when the spring is about to move(at x=0.5m)... 'cause i thought the friction force will be exerted after this point reducing the acceleration...
What's wrong in my thought?

And, for the second question, i tried to get the velocity when x=1m using energy-conservation theorem, and i got m*v² = k*x² and hence v = 3.53

Now i used the constant acceleration formula, the final velocity is zero, and initial v = 3.53,
a= $$\mu$$mg/2 and the displacement = 1.

But then i got the answer as 0.62 which is wrong...

What's wrong in here?


Please help me out with those tiny little(?) two questions...

 

[MiniSmSm]

I think it`s related to hook`s law ( F =-kx )
and Newton`s second law
hope somebody help us to solve this

 

[kerimek]

I would like to clarify that the maximum acceleration of the block does indeed occur at the instant it begins to move, at x=0.5m. This is because at this point, the block is experiencing the maximum force from the compressed spring, and there is no friction acting on it yet. Once the block starts moving, the friction force will start to oppose its motion, causing the acceleration to decrease.

For the second question, you are on the right track by using the energy conservation theorem. However, there are a few things to consider. First, when solving for the velocity at x=1m, you need to take into account the fact that the block is still in contact with the spring at this point, so the energy stored in the spring is still contributing to the block's velocity. So the correct equation would be m*v^2 = (1/2)k*x^2 + (1/2)mv^2, where x=1m. Solving for v, you should get v=2.83 m/s.

Next, when using the constant acceleration formula, you need to consider the total distance traveled by the block, which is 2m (from x=1m to x=2m). So the equation should be 0 = v^2 + 2a(2-1), where v=2.83 m/s and a is the acceleration due to friction. Solving for a, you should get a= 2.83 m/s^2.

Finally, you can use the equation F(friction) = μk*mg to solve for the coefficient of kinetic friction. Plugging in the values, you should get μk = 0.283. So the coefficient of kinetic friction for the surface is 0.283.

I hope this helps clarify the concepts and equations involved in solving this problem. Keep up the good work in your studies of kinematics!