When the speed booster is set up horizontally at the edge of a table (h=1.11 m) it sends a toy car of mass m=0.06kg in the projectile motion, so that it lands at horizontal distance of x1=1.36 m away from the edge of the table.
When the sam car with the same speed booster are used but the car is made to go through the loop-the-loop track before it is send away from the edge of the table, it lands
at distance x2=1.08.
Use this information to find the energy lost by the car while making the vertical loop
To find the energy lost by the car while making the vertical loop, we need to compare the initial kinetic energy (KE1) of the car when it launches horizontally from the edge of the table with the final kinetic energy (KE2) of the car when it lands after going through the loop-the-loop track.
First, let's calculate the initial kinetic energy (KE1) of the car when it launches horizontally. We can use the basic equation for kinetic energy:
KE1 = (1/2) * m * v1^2
where m is the mass of the car and v1 is the velocity of the car as it leaves the speed booster horizontally.
Next, we'll calculate the final kinetic energy (KE2) of the car when it lands at distance x2 after going through the loop-the-loop track. Since the car goes through a vertical loop, we know that the car loses some of its initial kinetic energy due to work done against gravity and friction.
To calculate KE2, we can use the conservation of mechanical energy, which states that the total mechanical energy remains constant in the absence of non-conservative forces like friction. In this case, we assume that the only non-conservative force is friction, and it causes the car to lose energy.
The mechanical energy consists of the sum of the kinetic energy (KE) and the potential energy (PE):
Mechanical Energy = KE + PE
Since we are only interested in the energy lost, we can write the equation as:
KE2 = KE1 - Energy Lost
Now, let's substitute the given values into the equations:
Mass of the car, m = 0.06 kg
Distance from the edge of the table, x1 = 1.36 m
Distance from the edge of the table after going through the loop, x2 = 1.08 m
Height of the table, h = 1.11 m
We need to find the velocity v1 of the car as it leaves the speed booster. To do this, we can use the equation of projectile motion for horizontal distance:
x = v1 * t
Since the car is launched horizontally, the initial vertical velocity is zero. Hence, we only consider the horizontal distance traveled by the car.
Substituting the values into the equation, we have:
x1 = v1 * t
1.36 m = v1 * t
Similarly, we can find the time t taken by the car to reach the edge of the table after going through the loop. Again, we only consider the horizontal distance traveled:
x2 = v2 * t
1.08 m = v2 * t
Now, we solve these two equations to get the values of v1 and v2:
v1 * t = 1.36 m (Equation 1)
v2 * t = 1.08 m (Equation 2)
Dividing Equation 1 by Equation 2:
(v1 * t) / (v2 * t) = (1.36 m) / (1.08 m)
v1 / v2 = 1.26
Now, we can use the conservation of mechanical energy to find the energy lost. Since the car starts at a height of 1.11 m and lands at a height of 0 m, we can write:
KE2 = KE1 - m * g * h
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Let's plug in the values:
KE2 = KE1 - (0.06 kg) * (9.8 m/s^2) * (1.11 m)
To find the energy lost, we can subtract KE2 from KE1:
Energy Lost = KE1 - KE2
Finally, to calculate the energy lost by the car while making the vertical loop, we need to know the value of the initial kinetic energy (KE1) of the car. This value can be calculated using the initial velocity (v1) obtained from dividing Equation 1 by Equation 2:
KE1 = (1/2) * m * v1^2
Plugging in the values, you can find the energy lost by the car while making the vertical loop.