Question Video: Finding the Speed of a Body Moving on a Smooth Surface under the Action of a Force | Nagwa Question Video: Finding the Speed of a Body Moving on a Smooth Surface under the Action of a Force | Nagwa

# Question Video: Finding the Speed of a Body Moving on a Smooth Surface under the Action of a Force Mathematics • Third Year of Secondary School

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A block of mass 5 kg is initially at rest on a smooth surface. A constant force of magnitude 39 N acting parallel to the surface is applied to the block. Determine the speed of the block 9 seconds after the force was initially applied.

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### Video Transcript

A block of mass five kilograms is initially at rest on a smooth surface. A constant force of magnitude 39 newtons acting parallel to the surface is applied to the block. Determine the speed of the block nine seconds after the force was initially applied.

Let’s begin by drawing a free-body diagram of this scenario. We have a block that is at rest on a horizontal surface. Since it’s at rest, we can say its initial velocity 𝑢, which is sometimes represented as 𝑣 sub zero, is equal to zero. It has a mass of five kilograms, which means it exerts a downwards force of its weight on the surface on which it rests. We can use Newton’s second law of motion to calculate the exact value of this force. Its force is equal to mass times acceleration. Well, the block itself has a mass of five kilograms, and the acceleration due to gravity we’ll say is 𝑔. And that’s equal to 9.8 meters per square second.

By Newton’s third law of motion, since the body itself exerts a downwards force on the surface, the surface must exert a force equal in magnitude and opposite in direction on the block. We’ll call that force 𝑅. Next, we’re told a constant force of magnitude 39 newtons acts parallel to the surface. And since the surface is smooth, this is the only force acting in this direction; there’s no frictional force. Now we want to find the speed of the block, let’s call that 𝑣, nine seconds after the force was initially applied. And so we have three values here. We have the initial speed is zero, the final velocity 𝑣 is what we’re trying to find, and 𝑡, time, is nine seconds.

Now, it follows that if we can find the value of 𝑎, acceleration, or 𝑠, displacement, we’ll be able to use one of our equations of constant acceleration to calculate 𝑣. In fact, we can use Newton’s second law of motion to calculate the acceleration of the block. Thinking about the forces that act on the block in the horizontal direction, we know that the sum of these forces is simply 39. And we’re taking the direction in which the block is moving to the right to be positive. The mass of the block is five, so mass times acceleration is five 𝑎. And we have an equation for 𝑎; it’s 39 equals five 𝑎. Dividing both sides of this equation by five and we find 𝑎 is equal to 39 over five meters per square second.

And so we now have everything we need to be able to apply one of our equations of constant acceleration. The equation that we’ll use is 𝑣 equals 𝑢 plus 𝑎𝑡. Since the initial speed is zero, the acceleration is 39 over five, and 𝑡 is nine, 𝑣 is zero plus 39 over five times nine, which is equal to 351 over five or equivalently 70.2.

The speed of the block then nine seconds after the force was initially applied is 70.2 meters per second.

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