A bloc of 10 kg is put at the top of an inclined plan of 45 degrees (to the left), attached to a spring which has a spring constant of 250 N/m. The coefficient of kinetic friction between the bloc and the surface of the inclined plan is of 0,300.
What is the maximum elongation of the spring?
I know that maximum elongation must mean that when the block has a velocity of 0 m/s. Are speeds applied to springs as well?
Huh?
the force down the plane is mg*sinTheta.
Now, the energy in the spring will be 1/2 kx^2,and the energy lost due to friction will be x*mg*cosTheta*mu
so force*x=1/2 kx^2=mgCosTheta*mu*x
that is , the work done by gravity equals the work done by friction+energyinspring
is it always sinTheta for a force down a plane? Or does it depend on the inclination (left, right)
and what does mu stand for in x*mg*cosTheta*mu ?
Thank you for your time!
mu stands for the coefficent of friction
Okay so from what I understand, I find the elongation from the total Work done on the spring and of course by isolating it in my equation. Sorry if my terms aren't good, I do it in french and try to translate it as good as possible
To find the maximum elongation of the spring, we need to consider the forces acting on the block at that point. Since the velocity of the block is zero, the net force acting on it must also be zero.
The forces acting on the block are:
1. The weight of the block, which acts vertically downwards with a magnitude of mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The force of friction acting on the block, which opposes its motion. This force can be calculated as μN, where μ is the coefficient of kinetic friction and N is the normal force (equal to the weight of the block in this case).
3. The force exerted by the spring, which is proportional to the elongation or compression of the spring. This force can be calculated as kx, where k is the spring constant and x is the elongation or compression of the spring.
At the maximum elongation, these forces balance each other out:
Weight of the block = Force of friction + Force exerted by the spring
mg = μN + kx
In this case, the normal force N is equal to the weight of the block, which is mg. Substituting this in the equation, we get:
mg = μmg + kx
Simplifying,
0 = μmg + kx - mg
0 = μmg - (1 - k/m)mg
0 = (μ - (1 - k/m))mg
For the block to be in equilibrium (with zero velocity), the expression in the brackets must be zero:
μ - (1 - k/m) = 0
Solving for μ, we get:
μ = 1 - k/m
Now we can substitute the given values:
μ = 1 - (250 N/m) / (10 kg)
μ = 1 - 25 N/m
μ = 0.300 (given)
0.300 = 1 - 25 N/m
25 N/m = 1 - 0.300
25 N/m = 0.700
Now we can solve for x, the maximum elongation of the spring:
kx = (25 N/m)(0.700)
x = (25 N/m)(0.700) / 250 N/m
x = 0.175 m
Therefore, the maximum elongation of the spring is 0.175 meters.