You are launching a 2-kg potato out of a potato cannon. The cannon is 1.5m long and is aimed 30 degrees above the horizontal. It exerts a 50N force on the potato. What is the kinetic energy of the potato as it leaves the muzzle of the potato cannon?
To calculate the kinetic energy of the potato as it leaves the muzzle of the potato cannon, you need to use the formulas for kinetic energy and work.
The formula for kinetic energy is:
KE = 1/2 * m * v^2
where KE is the kinetic energy, m is the mass of the potato, and v is the velocity of the potato.
The formula for work is:
W = F * d * cos(theta)
where W is the work done, F is the force exerted on the potato, d is the distance traveled by the potato, and theta is the angle between the force and the direction of motion.
In this case, the force F exerted on the potato is 50N, and the distance traveled by the potato is the length of the cannon, which is 1.5m. The angle theta is 30 degrees above the horizontal, so you need to use the cosine of 30 degrees.
First, let's find the work done on the potato:
W = F * d * cos(theta)
W = 50N * 1.5m * cos(30)
W = 50N * 1.5m * 0.866
W ≈ 64.95 Joules
The work done on the potato is 64.95 Joules.
Since the work done on the potato is equal to its kinetic energy, we can substitute this value into the formula to find the kinetic energy:
KE = 1/2 * m * v^2
64.95 = 1/2 * 2kg * v^2
64.95 = 1 * v^2
v^2 ≈ 64.95
v ≈ √64.95
v ≈ 8.07 m/s
Therefore, the kinetic energy of the potato as it leaves the muzzle of the potato cannon is approximately 64.95 Joules, and its velocity is approximately 8.07 m/s.