Video: Speed Increase of an Object Accelerated across a Distance

A cart with a mass of 18 kg is at rest on a level floor. A constant 15-N force is applied to the cart at an angle 52° below the horizontal. If friction is negligible, what is the speed of the cart when the force is applied over a distance 6.7 m?

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Video Transcript

A cart with a mass of 18 kilograms is at rest on a level floor. A constant 15-newton force is applied to the cart at an angle 52 degrees below the horizontal. If friction is negligible, what is the speed of the cart when the force is applied over a distance 6.7 meters?

In this exercise, we want to solve for the speed of the cart after the force has been applied over some distance. We’ll call that speed 𝑣 sub 𝑓. We’re told the cart’s mass, 18 kilograms, we’ll label that 𝑚, the force magnitude acting on the cart 15 newtons, which we’ll call 𝐹. The direction relative to the horizontal that that force acts 52 degrees, we’ll name that angle 𝜃, and finally, we’re told the distance over which that force 𝐹 is applied. It’s 6.7 meters. We’ll label that distance 𝑑.

Let’s start on our solution by drawing a diagram of the scenario. We start off in this example with our cart of mass 𝑚 at rest on a flat frictionless surface. A force 𝐹 is applied to the cart in the direction 𝜃 degrees below the horizontal. This force continues to act on the cart over a distance, 𝑑, of 6.7 meters. After being under the influence of the force over that span, we want to know what is the final speed of the cart.

To find out, let’s recall Newton’s second law of motion. The second law tells us that the net force that acts on an object is equal to that object’s acceleration times its mass. In our case, as we look at the force applied to our mass, we see that only a component of the force will contribute to its acceleration to the right. The vertical component of that force will contribute to the normal force that the mass experiences. Only the horizontal component of 𝐹 will contribute to its acceleration to the right.

Looking at our diagram, we see that it’s the cos of that angle 𝜃 multiplied by the magnitude of the force 𝐹 which is equal to the net force to the right on our case. By the second law, that’s equal to the mass of the case multiplied by its acceleration. If we divide both sides of this equation by mass 𝑚, canceling that out on the right-hand side, we see that the acceleration of our object is equal to 𝐹 times the cos of 𝜃 all divided by the object’s mass, 𝑚.

Since all these terms are constant, acceleration 𝑎 is also constant over the span of the distance 𝑑. When the acceleration of an object is constant, that’s a sign that the kinematic equations apply to describe the motion of that object. Looking over these four equations, we seek to find one which matches what we want to solve for, a final velocity, as well as the information we’re given in the problem statement.

The second equation we’ve written is a match for these conditions. It tells us that the final speed of the cart squared is equal to its initial speed squared plus two times its acceleration, again which is constant, multiplied by the distance over which that acceleration occurs. We’re told in the problem statement that our case starts at rest, which means 𝑣 sub zero is equal to zero. If we then take the square root of both sides of this equation, we see that 𝑣 sub 𝑓 is equal to the square root of two times 𝑎 times 𝑑. And 𝑎, the acceleration, is something we’ve solved for symbolically earlier.

If we plug in this expression for 𝑎, we now have an expression for what we want to solve for, 𝑣 sub 𝑓, in terms of known values 𝐹, 𝜃, 𝑚, and 𝑑. We’re ready to plug in and solve for 𝑣 sub 𝑓. With these values inserted, when we calculate the result of the square root, we find that, to two significant figures, it equals 2.6 meters per second. That’s our object’s final velocity after being accelerated over the distance 𝑑.

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