A child's toy consists of a toy car of mass 0.100kg which is able to roll without friction on the loop-the-loop track shown. The car is accelerated from rest by pushing it with force F over a distance d, then the car slides with a constant velocity until it encounters the loop. The radius of the loop is R.
A.The Child can push with a constant force of 3.20N. Use the Work-Kinetic Energy Theorem to find the speed of the car at the end of a 1.50m push.
B. If the radius of the loop is R=80.0cm what is the speed of the car when it reaches the top of the loop/
C. What is the minimum distance d that the child must push for the car to stay on the track at the top of the loop?
A. To find the speed of the car at the end of a 1.50m push, we can use the Work-Kinetic Energy Theorem, which states that the work done on an object is equal to its change in kinetic energy.
The work done on the car by the child pushing it is given by the product of the force and the distance pushed: Work = Force x Distance. In this case, the force is 3.20N and the distance is 1.50m.
Next, we need to calculate the change in kinetic energy of the car. Initially, the car is at rest, so its initial kinetic energy is zero. At the end of the push, the car has gained kinetic energy, which can be calculated using the work done on it. So, we have:
Work = Change in Kinetic Energy
Solving for the change in kinetic energy, we get:
Change in Kinetic Energy = Work
Now, we can plug in the values:
Change in Kinetic Energy = 3.20N x 1.50m
After calculating the change in kinetic energy, we can convert it to the speed of the car using the formula for kinetic energy:
Kinetic Energy = (1/2) x mass x speed^2
Rearranging the equation, we get:
Speed = sqrt((2 x Kinetic Energy) / mass)
Finally, we can substitute the calculated change in kinetic energy and the given mass (0.100kg) into the equation to find the speed of the car.
B. To find the speed of the car when it reaches the top of the loop, we can use the principle of conservation of mechanical energy. At the top of the loop, the car has both kinetic and potential energy.
The total mechanical energy at the top of the loop is equal to the sum of the car's kinetic energy and potential energy. The kinetic energy can be calculated using the formula mentioned earlier, and the potential energy can be calculated using the formula:
Potential Energy = mass x g x height
In this case, the height is the radius of the loop (R), and g is the acceleration due to gravity.
Since mechanical energy is conserved, the total mechanical energy at the top of the loop is equal to the initial mechanical energy at the end of the push. So, we can equate the initial kinetic energy (calculated in part A) to the total mechanical energy at the top of the loop.
Solving for the speed at the top of the loop, we can use the same formula as in part A to calculate it.
C. To find the minimum distance (d) that the child must push for the car to stay on the track at the top of the loop, we need to consider the forces acting on the car at that point.
At the top of the loop, the car is in contact with the track, experiencing both the force of gravity and the normal force from the track. The net force is the difference between these two forces, and it must provide the necessary centripetal force to keep the car moving in a circular path without falling off the track.
The minimum distance (d) is the distance required for the car to reach the top of the loop with sufficient speed so that the net force can provide the necessary centripetal force.
Using Newton's second law, we can equate the net force to the mass times centripetal acceleration:
Net Force = mass x centripetal acceleration
The centripetal acceleration can be calculated using the formula:
Centripetal Acceleration = (velocity^2) / radius
Solving for the velocity, we can use the same formula as in part B to calculate it.
Substituting the calculated velocity and the given radius (R) into the equation for the net force, we can find the minimum distance (d) required.