Question

You are launching a 2 kg potato out of a potato cannon. The cannon is 1.1 m long and is aimed 65 degrees above the horizontal. It exerts a 48 N force on the potato. What is the kinetic energy (in J) of the potato as it leaves the muzzle of the potato cannon?

Answer 1

Answer:51.84

Explanation:

Given

cannon is 1.1 m long

Force on potato is 48 N

angle of launch $\theta =65^{\circ}$

Net force at the end of cannon

65-mgsin(65)=65-17.78=47.21 N

acceleration is given by$=\frac{47.21}{2}=23.60 m/s^2$

Velocity acquired during travel of 1.1 m

$v^2-u^2=2as$

$v^2=2\times 23.60\times 1.1$

v=7.20 m/s

kinetic energy of the potato $=\frac{mv^2}{2}$

$K.E.=\frac{2\times 7.20^2}{2}=51.84 J$

Answer 2

The kinetic energy of the potato as it leaves the muzzle of the potato cannon is 29.53 J

What is kinetic energy?

Kinetic energy is the energy of a body in motion.

To calculate the kinetic energy of the potato as it leaves the muzzle, we use the formula below.

Formula:

• K.E = Fdcos∅............ Equation 1

Where:

• K.E = Kinetic energy of the potato
• F = Force exerted on the potato
• d = Length of the cannon
• ∅ = angle of the cannon to the horizontal

From the question,

Given:

• F = 48 N
• d = 1.1 m
• ∅ = 65°

Substitute these values into equation 1

• K.E = 48×1.1×cos56°
• K.E = 29.53 J.

Hence, The kinetic energy of the potato as it leaves the muzzle of the potato cannon is 29.53 J.

Learn more about kinetic energy here: https://brainly.com/question/25959744