Please explain the answer! A block with mass m = 13 kg rests on a frictionless t
ID: 1344152 • Letter: P
Question
Please explain the answer!
A block with mass m = 13 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 5023 N/m after being compressed a distance x1 = 0.424 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.2 m long. For this rough path, the coefficient of friction is k = 0.47.
1) Instead, the spring is only compressed a distance x2 = 0.103 m before being released. How far into the rough path does the block slide before coming to rest?
2) What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released?
Explanation / Answer
1)
Use conservation of energy to determine the energy of the block as it leaves the spring. The potential energy of the spring becomes the kinetic energy of the block.
As for a spring, F=kx, where F is force, k is the spring constant, and x is the distance the spring is stressed or compressed from equilibrium, U (potential energy) = .5(x^2), as energy is the integral of velocity with respect to distance (x). K is kinetic energy.
K=U=.5*x^2
K=.5*(5023 N/m)*(.103 m)^2 = 26.64 J
We can use this energy value to determine how far the block will go as the friction will "consume" all of this kinetic energy.
Now as you know, the magnitude of the force of kinetic friction on a flat surface is equivalent to...
F=mg
Now the integral of force with respect to displacement (or force * displacement) is work, so...
W=mgy
Where y is your answer...so if we set K=W...
K=mgy
K/(mg)=y
(26.64 J)/((0.47)*(13 kg)*(9.8 m/s^2)) = y = .445 m
2) What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released
The work done by the spring must be "barely" greater than than the worjk done by the friction!
0.5 * k * x^2 = u * m* g * d
x^2 = (u * m* g * d) / (0.5 * k) = (0.47 * 13 * 9.8 * 0.445) / (0.5 * 5023) = 0.0106
x = 0.1029 m
If the the spring is compressed barely more than 0.1029 meter, the block will come to rest at a point barely past the 0.445 m long rough patch!
If the rough patch is the original 2.2 m long:
x^2 = (u * m* g * d) / (0.5 * k) = (0.47 * 13 * 9.8 * 2.2) / (0.5 * 5023) =
x = 0.229 m