Please anybody that can help to understand this exercises 1.- a)Drop a one kilog
ID: 1453597 • Letter: P
Question
Please anybody that can help to understand this exercises 1.- a)Drop a one kilogram mass from height of 2 meters onto a vertically oriented Spring( spring constant k=100) Find the maximum compression distance, and the compression at equilibrium when the mass rest.
b) Now orient the spring horizontally on a flat and level table top. Fix one end of the spring to a wall. Compress the other end with the one kilogram mass by 25 cm. How far will the mass slide on the table with mu(static)= 0.4 and mu (kinetic)= 0.35? The mass and the spring are not connected, and the mass of the spring is negligible. What is the total work done by friction?
Please anybody that can help to understand this exercises 1.- a)Drop a one kilogram mass from height of 2 meters onto a vertically oriented Spring( spring constant k=100) Find the maximum compression distance, and the compression at equilibrium when the mass rest.
b) Now orient the spring horizontally on a flat and level table top. Fix one end of the spring to a wall. Compress the other end with the one kilogram mass by 25 cm. How far will the mass slide on the table with mu(static)= 0.4 and mu (kinetic)= 0.35? The mass and the spring are not connected, and the mass of the spring is negligible. What is the total work done by friction?
Please anybody that can help to understand this exercises
b) Now orient the spring horizontally on a flat and level table top. Fix one end of the spring to a wall. Compress the other end with the one kilogram mass by 25 cm. How far will the mass slide on the table with mu(static)= 0.4 and mu (kinetic)= 0.35? The mass and the spring are not connected, and the mass of the spring is negligible. What is the total work done by friction?
Explanation / Answer
1.
a)
When the mass is dropped from a height h, it will have a poetential energy mgh. This potential energy is converted into the energy of the spring.
mgh = 1/2 kx2
x2 = 2mgh/k = (2 x 1 x 9.8 x 2) / 100
x = sqrt(0.392) = 0.626 m
When it reaches an equilibrium, downward gravitational force on the mass equal the spring force
mg = kx
x = mg/k = (1 x 9.8) / 100 = 0.098 m
b)
The energy stored in the spring will be converted into the kinetic energy of the mass.
1/2 kx2 = 1/2 mv2.
1/2 mv2 = 0.5 x 100 x (0.25)2
= 3.125 J
Frictional force, which acts against the moving mass will make it stop after doing a work
Frictional force = mg
= 0.35 (coefficient of kinetic friction)
Work done by frictional force = mg x distance
mg x distance = 3.125 J
0.35 x 1 x 9.8 x distance = 3.125
distance = 0.911 m
Work done by frictional force = 3.125 J