The CoolWheels Toy Company is designing a new track where a car (mass 60 g) is launched by a spring-loaded starter (k=11 N/m) and goes through a vertical loop (radius 9 cm). If the spring is depressed ∆x=19 cm from equilibrium, what is the speed of the car at the top of the loop?
initial PEspring-changePE top= final KE
1/2 11*.19^2 - .060g*.18=1/2*.060v^2 solve for v.
Thank you
To find the speed of the car at the top of the loop, we need to consider the conservation of mechanical energy. The mechanical energy of the car is the sum of its kinetic energy and potential energy.
First, let's calculate the potential energy of the spring when it is depressed ∆x. The potential energy stored in a spring is given by the equation:
Potential Energy = (1/2) * k * (∆x)^2,
where k is the spring constant and ∆x is the displacement of the spring from its equilibrium position.
Plugging in the given values:
Potential Energy = (1/2) * 11 N/m * (0.19 m)^2
= (1/2) * 11 N/m * 0.0361 m^2
= 0.19855 J
Now, let's calculate the total mechanical energy of the car when it is at the bottom of the loop. At this point, the car has no gravitational potential energy and only has kinetic energy. The kinetic energy is given by the equation:
Kinetic Energy = (1/2) * m * v^2,
where m is the mass of the car and v is its speed.
Plugging in the values:
Kinetic Energy = (1/2) * 0.060 kg * v^2
Since the total mechanical energy is conserved, the sum of potential energy and kinetic energy at the bottom of the loop is equal to the kinetic energy at the top of the loop. Therefore:
Potential Energy (at bottom) + Kinetic Energy (at bottom) = Kinetic Energy (at top)
0 + (1/2) * 0.060 kg * v^2 = Kinetic Energy (at top)
Now, let's calculate the kinetic energy at the top of the loop. At this point, the car is at its maximum height and has no kinetic energy. Its energy is entirely potential energy, given by the equation:
Potential Energy = m * g * h,
where m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the loop (which is equal to twice the radius of the loop).
Plugging in the values:
Potential Energy (at top) = 0.060 kg * 9.8 m/s^2 * (2 * 0.09 m)
= 0.1062 J
Since the total mechanical energy is conserved, we can equate the potential energy at the top with the kinetic energy at the bottom:
0.1062 J = (1/2) * 0.060 kg * v^2
Now, we can solve for v:
0.1062 J = (1/2) * 0.060 kg * v^2
Simplifying and rearranging the equation:
v^2 = (2 * 0.1062 J) / 0.060 kg
v^2 = 3.54 m^2/s^2
Taking the square root of both sides to solve for v:
v = √(3.54 m^2/s^2)
v ≈ 1.88 m/s
Therefore, the speed of the car at the top of the loop is approximately 1.88 m/s.