a 21.6 gram arrow is shot through a 7.5 cm apple. if the arrow enters the apple at 35.5 m/s and emerges with a speed of 25.3 m/s in the same direction, what is the magnitude of the force with which the apple has resisted the arrow? (assume force is exerted only on the tip of the arrow)



Answer :

The magnitude of the force with which the apple has resisted the arrow is 88.06 Newton.

Using, work-energy theorem,

According to which, the work done by the all the forces on the body is equal to the change in kinetic energy of the body.

So, we can write,

Work done = ∆KE

We know,

Work done = force × displacement

The force is the resistive force by the apple and the displayed is given to be 0.075 meters.

So, we can write,

F(0.075) = 1/2M(V²-U²)

Where,

M is the mass of apple,

V is final velocity,

U is initial velocity.

Putting all the values,

F(0.075) = 1/2(0.0213)((25.3)²-(35.5)²)

Solving further,

F = 88.06 Newton

The resistive force of the apple is 88.06 Newton.

To know more about Work-Energy theorem, visit,

https://brainly.com/question/22236101

#SPJ4