A cart of mass 223 g is placed on a frictionless horizontal air track. A spring having a spring constant of 10.20 N/ m is attached between the cart and the left end of the track. When in equilibrium, the cart is located 11.0 cm from the left end of the track. If the cart is displaced 4.20 cm from its equilibrium position, find
(c) its speed when it is 14.0 cm from the left end of the track.
When the cart is 14 cm from the left end of the track, it has a displacement of
x=14 -11 cm =3 cm= 0.03 m from the equilibrium position.
The speed of the cart at this distance from equilibrium is
v=sqrt{k(A²-x²)m}=…
ok, it released spring energy 0.2cm from the starting position to the mark.
energyreleased=INT force*dx from 4.2 to 4.0 cm if you are a calculus person.
for non calculcus persons the KE it has is the difference of spring PE
KE=energyat4.2-energyat4.0
= 1/2 k (4.2^2 -4^2)
then knowing KE, solve for velocity
To find the speed of the cart when it is 14.0 cm from the left end of the track, we'll need to use the principles of energy conservation.
1. Calculate the potential energy at the equilibrium position:
The potential energy stored in a spring is given by the formula:
PE = (1/2)kx²
Where:
k = spring constant = 10.20 N/m
x = equilibrium position = 11.0 cm = 0.11 m
Substituting the values into the formula:
PE = (1/2)(10.20 N/m)(0.11 m)²
PE = 0.060105 J
2. Calculate the potential energy at 14.0 cm from the left end of the track:
x = displacement from equilibrium position = 14.0 cm = 0.14 m
PE = (1/2)kx²
PE = (1/2)(10.20 N/m)(0.14 m)²
PE = 0.127176 J
3. Calculate the kinetic energy:
The total mechanical energy (E) of the system is conserved, which means the sum of the potential and kinetic energy remains constant. Thus, when the potential energy changes, the kinetic energy will change to compensate.
KE = E - PE
Substituting the values:
KE = 0.060105 J - 0.127176 J
KE = -0.067071 J
The negative value indicates that the kinetic energy is less than the potential energy. It means the cart will have slower speed at 14.0 cm from the left end of the track compared to the equilibrium position.
4. Calculate the speed:
To find the speed, we'll use the formula for kinetic energy:
KE = (1/2)mv²
Where:
m = mass of the cart = 223 g = 0.223 kg
v = speed of the cart
Rearranging the formula:
v² = (2KE) / m
v² = (2)(-0.067071 J) / 0.223 kg
v² = -0.6029 J / kg
Taking the square root to find the speed:
v = √(-0.6029 J / kg)
v ≈ 0.617 m/s
Therefore, the speed of the cart when it is 14.0 cm from the left end of the track is approximately 0.617 m/s.
To find the speed of the cart when it is 14.0 cm from the left end of the track, we need to use the principles of conservation of energy.
First, let's find the potential energy of the spring when the cart is 11.0 cm from the left end (equilibrium position). The potential energy of a spring is given by the formula:
PE = (1/2)kx^2
where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
Substituting the given values, we have:
PE = (1/2)(10.20 N/m)(0.11 m)^2
= 0.06217 J
Next, let's find the potential energy of the spring when the cart is 14.0 cm from the left end. Using the same formula, we have:
PE = (1/2)(10.20 N/m)(0.14 m)^2
= 0.10164 J
Now, since the air track is frictionless, the total mechanical energy of the system (cart + spring) remains constant. Therefore, the sum of the kinetic energy and potential energy at any point will be equal to the initial potential energy at the equilibrium position.
At the equilibrium position, all the potential energy is stored in the spring, so the initial potential energy is 0.06217 J.
When the cart is 14.0 cm from the left end, all the potential energy is converted into kinetic energy. So,
KE = PE (at 14.0 cm)
= 0.10164 J
The kinetic energy is given by the formula:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the cart, and v is the velocity (speed) of the cart.
We know the mass of the cart is 223 g, which is 0.223 kg.
Substituting the values, we have:
0.10164 J = (1/2)(0.223 kg)v^2
Simplifying the equation:
0.20328 = 0.1115v^2
Dividing both sides of the equation by 0.1115:
v^2 = 1.825
Taking the square root of both sides of the equation:
v = sqrt(1.825)
= 1.35 m/s
Therefore, the speed of the cart when it is 14.0 cm from the left end of the track is approximately 1.35 m/s.