In Figure 8-34, a 1.2 kg block is held at rest against a spring with a force constant k = 785 N/m. Initially, the spring is compressed a distance d. When the block is released, it slides across a surface that is frictionless, except for a rough patch of width 5.0 cm that has a coefficient of kinetic friction µk = 0.44. Find d such that the block's speed after crossing the rough patch is 2.0 m/s.
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To find the value of d, we need to consider the forces acting on the block and apply Newton's laws of motion.
Let's start by analyzing the forces present in the system. The main forces at play are the force exerted by the spring (Fs) and the force of kinetic friction (Ff) on the rough patch.
1. Force exerted by the spring (Fs):
The force exerted by the spring is given by Hooke's Law:
Fs = -k * x,
where k is the force constant of the spring and x is the displacement of the spring from its equilibrium position.
2. Force of kinetic friction (Ff):
The force of friction can be calculated using the formula:
Ff = µk * Fn,
where µk is the coefficient of kinetic friction and Fn is the normal force exerted on the block.
Let's proceed with the solution:
Step 1: Find the normal force
The normal force is the force exerted by the surface on the block and is equal to the gravitational force acting on the block when the block is at rest:
Fn = m * g,
where m is the mass of the block and g is the acceleration due to gravity.
Given that the mass of the block is 1.2 kg, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate Fn as:
Fn = 1.2 kg * 9.8 m/s^2 = 11.76 N.
Step 2: Calculate the force of friction
Using the formula for the force of friction, Ff = µk * Fn, we can substitute the given values:
Ff = 0.44 * 11.76 N = 5.17 N.
Step 3: Determine the maximum distance for the block to achieve the desired speed
Since there is no friction on the surface except for the rough patch, the block will experience a net force only over this distance. To achieve a final speed of 2.0 m/s, we can use kinematic equations. Let's consider the final velocity (vf) as the unknown variable.
Using the equation:
vf^2 = vi^2 + 2 * a * d,
where vi = 0 (initial velocity), a = Fnet / m (acceleration), and d is the distance.
Since Fnet = Ff and the mass of the block is 1.2 kg, we can write:
vf^2 = 0 + (Ff / m) * d,
(2 m^2/s^2) = (5.17 N / 1.2 kg) * d,
1.67 m^2/s^2 = 4.31 N/kg * d.
Solving for d, we get:
d = (1.67 m^2/s^2) / (4.31 N/kg) = 0.387 m.
Therefore, the value of d such that the block's speed after crossing the rough patch is 2.0 m/s is approximately 0.387 m.