A 1.07 kg block slides across a horizontal surface directly toward a massless spring with spring constant 4,794 N/m. The surface is frictionless except for a rough patch of length 0.461 m that has coefficient of kinetic friction 0.358. The initial velocity of the block is 4.12 m/s directed straight toward the spring. What is the maximum compression of the spring?
KE – W(fr) =PE(spring)
m•v²/2 - μ•m•g•s = k•x²/2
x= sqrt[(2/k) •(m•v²/2 - μ•m•g•s)] =
=5.5•10^-2 m =5.5 cm
To find the maximum compression of the spring, we need to analyze the motion of the block as it slides across the horizontal surface and comes to a stop at the spring.
First, let's determine the net force acting on the block when it is sliding on the frictionless surface. Since the surface is frictionless, the only force acting on the block is the force of gravity, mg, where m is the mass of the block (1.07 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Next, let's determine the acceleration of the block on the rough patch of length 0.461 m that has a coefficient of kinetic friction μ = 0.358. The frictional force can be calculated as μ times the normal force, where the normal force is equal to the weight of the block (mg). So, the frictional force is μmg.
The net force acting on the block on the rough patch is the difference between the force of gravity and the frictional force: Net force = mg - μmg = (1 - μ)m g.
Using Newton's second law of motion, F = ma, where F is the net force, m is the mass, and a is the acceleration, we can find the acceleration of the block on the rough patch.
(1 - μ)mg = ma
Rearranging the equation, we get:
a = (1 - μ)g
Now that we have the acceleration, we can calculate the time it takes for the block to come to a stop. Since the initial velocity of the block is 4.12 m/s and the final velocity is 0 m/s (due to stopping at the spring), we can use the kinematic equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
0 = 4.12 + a * t
Rearranging the equation, we get:
t = -4.12 / a
Now, let's calculate the distance the block travels on the rough patch using the formula:
s = ut + (1/2)at^2
where s is the distance traveled.
Substituting the values, we get:
s = 4.12 * t + (1/2) * a * t^2
Finally, we can calculate the maximum compression of the spring using Hooke's Law, which states that the force exerted by a spring is proportional to its displacement:
F = -kx
where F is the force applied by the spring, k is the spring constant, and x is the displacement (maximum compression of the spring).
Since the only force acting on the block when it comes to a stop is the force exerted by the spring, we have:
kx = (1/2) * mv^2
Simplifying the equation, we find:
x = (1/2) * (mv^2) / k
Substituting the given values, we can solve for x, which will give us the maximum compression of the spring.