In Figure 8-34, a 1.2 kg block is held at rest against a spring with a force constant k = 695 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 1.5 m/s.
energy stored in spring= 1/2 k d^2
energy in f riction= mg*mu*.05m
leftover energy= 1/2 m 1.5^2
well, initial energy=finalenergy+frictionwork
solve for d
energy in f riction= mg*mu*.05m = 5.886
leftover energy= 1/2 m 1.5^2 = 1.35
So, If I add the leftover energy and the energy in friction I get the initial energy.
Then what??
To find the distance d, we can use the principle of conservation of mechanical energy, which states that the initial mechanical energy of the system is equal to the final mechanical energy.
1. Determine the initial mechanical energy of the system:
The initial mechanical energy includes the potential energy stored in the compressed spring and the potential energy due to gravity.
The potential energy stored in the compressed spring is given by Us = (1/2)kx², where x is the displacement of the spring.
The potential energy due to gravity is given by Ug = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above a reference point (which we can assume to be zero).
2. Determine the final mechanical energy of the system:
The final mechanical energy includes the kinetic energy of the block after crossing the rough patch.
The kinetic energy of the block is given by Kf = (1/2)mv², where m is the mass and v is the speed.
3. Calculate the distance d:
a. Set up the equation for the initial mechanical energy:
Initial mechanical energy = Us + Ug
b. Set up the equation for the final mechanical energy:
Final mechanical energy = Kf
c. The initial mechanical energy is equal to the final mechanical energy:
Us + Ug = Kf
d. Substitute the expressions for the different energies:
(1/2)kx² + mgh = (1/2)mv²
e. Rearrange the equation to isolate the distance d:
(1/2)kx² + mgh = (1/2)mv²
(1/2)kx² = (1/2)mv² - mgh
(1/2)kx² = (1/2)mv² - mgd
f. Solve for d:
mgd = (1/2)mv² - (1/2)kx²
d = ((1/2)m(v² - 2kx²)) / mg
g. Substitute the given values:
m = 1.2 kg
k = 695 N/m
v = 1.5 m/s
x = ?
g = 9.8 m/s²
h. Solve for d:
d = ((1/2)(1.2 kg)(1.5 m/s)² - 2(695 N/m)x²) / (1.2 kg)(9.8 m/s²)
To find the value of d, we need to consider the forces acting on the block at different points during its motion. Let's break down the problem step by step:
Step 1: Initial Compression
Initially, when the block is held at rest against the spring, the spring exerts a force in the opposite direction. According to Hooke's law, the force exerted by the spring is given by F = k * x, where F is the force, k is the spring constant, and x is the compression distance.
Given that k = 695 N/m and the mass of the block is 1.2 kg, we can calculate the force exerted by the spring. However, the given values don't provide enough information to calculate the compression distance (d) at this point.
Step 2: Release of the Block
When the block is released, it starts moving and slides across the surface. Since it is a frictionless surface, there is no horizontal force acting on the block initially. The only force acting on it is the force exerted by the spring, which causes the block to move.
Step 3: Crossing the Rough Patch
As the block slides across the rough patch, the equation for the force of kinetic friction comes into play: fk = µk * N, where fk is the force of kinetic friction, µk is the coefficient of kinetic friction, and N is the normal force.
The normal force is equal to the gravitational force acting on the block, which is given by mg, where m is the mass of the block and g is the acceleration due to gravity.
Since the block is moving at a constant speed, the force of kinetic friction fk is equal to the force exerted by the spring, k * d. Therefore:
µk * N = k * d
Substituting the values provided in the problem, we have:
0.44 * (1.2 kg * 9.8 m/s^2) = 695 N/m * d
Simplifying the equation:
d ≈ (0.44 * 1.2 kg * 9.8 m/s^2) / 695 N/m
d ≈ 0.689 m
So, approximately, the value of d such that the block's speed after crossing the rough patch is 1.5 m/s is 0.689 m.