a green bead of mass 70 g slides along a straight wire. The length of the wire from point "A" to point "B" is 0.700 m, and point "A" is 0.400 m higher than point "B". A constant friction force of magnitude 0.020 0 N acts on the bead. If the bead is released from rest at point "A", what is its speed at point "B" ?
To find the speed of the bead at point B, we can use the principle of conservation of mechanical energy. The total mechanical energy of the bead at point A (when it is released from rest) is equal to the mechanical energy at point B.
The mechanical energy at any point is the sum of the kinetic energy and potential energy.
Let's calculate the potential energy at point A and point B:
Potential energy at point A:
PE(A) = m * g * h
Where:
m = mass of the bead = 70 g = 0.07 kg
g = acceleration due to gravity = 9.8 m/s²
h = height from point B to point A = 0.400 m
PE(A) = 0.07 kg * 9.8 m/s² * 0.400 m
PE(A) = 0.2744 J
Potential energy at point B:
PE(B) = m * g * h
Where:
h = 0 (as the bead is at ground level)
PE(B) = 0
Since the friction force is constant, it does not change the mechanical energy. So, the total mechanical energy at point A is equal to the total mechanical energy at point B.
Total Mechanical Energy at point A = Total Mechanical Energy at point B
Kinetic Energy at point A + Potential Energy at point A = Kinetic Energy at point B + Potential Energy at point B
The bead is released from rest at point A, so its initial velocity is zero. Hence, the kinetic energy at point A is zero.
0 + 0.2744 J = Kinetic Energy at point B + 0
Kinetic Energy at point B = 0.2744 J
The formula for kinetic energy is:
Kinetic Energy = (1/2) * m * v²
Where:
m = mass of the bead = 0.07 kg (given)
v = velocity at point B (to be found)
0.2744 J = (1/2) * 0.07 kg * v²
Now, we can solve for v:
v² = (2 * 0.2744 J) / 0.07 kg
v² = 7.8429 J / 0.07 kg
v² = 112.0429 m²/s²
Taking the square root of both sides:
v = √(112.0429 m²/s²)
v ≈ 10.591 m/s
Therefore, the speed of the bead at point B is approximately 10.591 m/s.
To find the speed of the bead at point B, we can use the principle of conservation of energy.
First, let's calculate the potential energy at point A:
Potential Energy (PE) = mass (m) * gravity (g) * height (h)
PE = 0.07 kg * 9.8 m/s^2 * 0.4 m
PE = 0.2744 J
Next, let's calculate the potential energy at point B:
PE = 0.07 kg * 9.8 m/s^2 * 0.7 m
PE = 0.4806 J
Since the bead is sliding along the wire, the mechanical energy is conserved. Therefore, the change in potential energy is equal to the change in kinetic energy.
Change in PE = -0.2062 J (0.4806 J - 0.2744 J)
Next, let's calculate the kinetic energy at point B:
Kinetic Energy (KE) = 0.5 * mass (m) * velocity (v)^2
Since the bead is released from rest, its initial velocity is 0.
KE = 0.5 * 0.07 kg * v^2
KE = 0.035 kg * v^2
Now, equating the change in potential energy to the kinetic energy at point B:
-0.2062 J = 0.035 kg * v^2
Solving for v^2:
v^2 = -0.2062 J / 0.035 kg
v^2 = -5.89 m^2/s^2
Since speed cannot be negative, we take the positive square root:
v = √(5.89 m^2/s^2)
v ≈ 2.4 m/s
Therefore, the speed of the bead at point B is approximately 2.4 m/s.