Video: Calculating the Force Applied That Brings a Moving Object to Rest over a Distance

A bullet has a mass of 2.60 g and moves horizontally at a speed of 335 m/s as it collides with a stack of eight pine boards, each 0.750′ thick. The bullet decelerates as it penetrates the boards. And the bullet comes to rest just as it has moved the full distance through all eight boards. Find the average force exerted by the boards on the bullet. Assume that the motion of the boards due to the bullet’s impact is negligible.

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Video Transcript

A bullet has a mass of 2.60 grams and moves horizontally at a speed of 335 meters per second as it collides with a stack of eight pine boards, each 0.750 inches thick. The bullet decelerates as it penetrates the boards. And the bullet comes to rest just as it has moved the full distance through all eight boards. Find the average force exerted by the boards on the bullet. Assume that the motion of the boards due to the bullet’s impact is negligible.

Looking to solve here for the average force exerted by the boards on a bullet, we can call that force 𝐹. And we’ll start on our solution by drawing a diagram of this situation. We have in this scenario a bullet with an initial speed 𝑣 sub 𝑖 of 335 meters per second encountering a stack of eight pine boards. Each of the boards has a thickness 𝑇 of 0.750 inches. And we’re told that by the time the bullet makes it all the way through the last board, it comes to a stop.

Knowing all this, we want to solve for the average force 𝐹 that the boards exert on the bullet as it decelerates. Seeking to solve for average force may remind us of Newton’s second law of motion, which says that the net force on an object equals its mass times its acceleration 𝑎. Using this law, we want to solve for 𝐹. And we’re given 𝑚 in the problem statement. But we don’t yet know the acceleration of the bullet as it comes to a stop. However, if we assume that acceleration 𝑎 is constant, then that means the kinematic equations apply for describing the motion of the decelerating bullet.

Looking over these four equations of motion, we see that the second one helps us solve for what we want to know — acceleration — in terms of values we’re given. Written in terms of the variables for our particular situation, we can say that the final speed of the bullet squared is equal to its initial speed squared plus two times its acceleration times the distance it travels eight times the thickness of a board.

We know the bullet ends up at rest. So 𝑣 sub 𝑓 is equal to zero. So when we rearrange to solve for 𝑎, it equals negative 𝑣 sub 𝑖 squared all over two times eight 𝑇. We can substitute this expression for acceleration in for 𝑎 in our equation for force. And we understand that even though the minus sign implies that the bullet is decelerating, which is true. Since we want to solve for the average force the boards exert on the bullet, which will be positive, we’ll change that to a plus sign.

Looking at this expression, we were given the mass of the bullet 𝑚 in the problem statement as well as the initial speed of the bullet 𝑣 sub 𝑖. We’re also told the thickness 𝑇 of the boards. But that thickness is currently expressed in inches and we like to convert it to meters. 0.750 inches is approximately 1.905 centimeters. When we plug our values into this expression, we’re careful to use a mass in units of kilograms and a distance of the thickness of each board in units of meters. Entering this expression on our calculator, it comes out to 960 newtons. To two significant figures, that’s the average force that the boards exert on the bullet.

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