You are launching a 2 kg potato out of a potato cannon. The cannon is 2.0 m long and is aimed 70 degrees above the horizontal. It exerts a 52 N force on the potato. What is the kinetic energy (in J) of the potato as it leaves the muzzle of the potato cannon?

The net force on the potatoes is given by:

F = 52 - mgSintheta

F = 52 - (2*9.8 * Sin70°)

F = 52 - 18.4

F = 33.58N

Using Newton's 2nd law

F = ma

a=F/m = 33.58 / 2 = 16.79m/s^2

Using the equation of motion:

V^2 = u^2 + 2as

V^2 = 0 + 2 * 16.79 x2

V^2 = 67.16

V=sqrt (68.16)

V = 8.195m/s This is the exit velocity of the potatoes

Kinetic energy, K. E = 1/2mv^2

KE = 1/2 * 2 * 8.195^2

KE = 67.16J