# Video: Finding the Velocity of a Body Moving on a Smooth Horizontal Plane Where a Force Acts on It

A body of mass 30.6 kg was at rest on a smooth horizontal plane. A horizontal force 51 N acted on it for 5 seconds. Determine the body’s speed at the end of the 5 seconds.

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### Video Transcript

A body of mass 30.6 kilograms was at rest on a smooth horizontal plane. A horizontal force 51 newtons acted on it for five seconds. Determine the body’s speed at the end of the five seconds.

Our first step here is to use Newton’s second law to calculate the acceleration. Newton’s second law states that force equals mass multiplied by acceleration, or 𝐹 equals 𝑚𝑎. We know that the mass is equal to 30.6 kilograms and the force is equal to 51 newtons. Substituting in these values gives us 51 is equal to 30.6𝑎. Dividing both sides by 30.6 gives us 𝑎 is equal to five over three, or five-thirds. We can, therefore, say that the acceleration of the body is five-thirds meters per second squared.

The question asks us to calculate the speed at the end of the five seconds. To answer this, we will use the equations of motion, often called the SUVAT equations. 𝑠 stands for the displacement. 𝑢 is the initial velocity. And 𝑣 is the final velocity. 𝑎 is the acceleration. And finally, 𝑡 is the time.

In this question, the body starts at rest. Therefore, 𝑢 is equal to zero. We are trying to calculate the final velocity 𝑣. 𝑎, as we have just worked out, is equal to five-thirds. And 𝑡 is equal to five. In this particular question, we are given no information about the displacement. The only one of the equations that doesn’t contain 𝑠 is 𝑣 equals 𝑢 plus 𝑎𝑡. Substituting in our values gives us 𝑣 is equal to zero plus five-thirds multiplied by five. This gives us a final velocity after five seconds of 25 over three, or twenty-five thirds meters per second.