High School

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?

Answer :

Answer:

Explanation:

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

The kinetic energy of the potato as it exits the potato cannon is 104 J, calculated by multiplying the force exerted by the cannon (52 N) with the displacement of the potato inside the cannon (2.0 m).

The kinetic energy of the potato as it leaves the muzzle of the potato cannon can be found by first calculating the work done on the potato by the cannon. Work is defined as the product of the force exerted on the object and the displacement in the direction of the force. Using the formula Work = Force x Displacement, we can calculate the work done by the cannon on the potato.

In this case, the force exerted by the cannon on the potato is 52 N and the displacement along the barrel of the cannon is 2.0 m, thus the work done is Work = 52 N x 2.0 m = 104 J. Since no other forces are doing work on the potato (friction is ignored), the work done on the potato is equal to its kinetic energy as it leaves the muzzle.

Therefore, the kinetic energy of the potato at the moment it exits the cannon is 104 Joules (J).

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