A 20.0 kg block is placed on a smooth horizontal table. A horizontal force of 12.0 Newton is applied to the block. The position of the block at 5.0 sec is
7.5 meter
Block Position Calculation Explained
This problem involves understanding the relationship between force, mass, acceleration, and displacement. We are given the mass of a block, a horizontal force applied to it, and the time for which the force acts. We need to find the final position (displacement) of the block, assuming it starts from rest on a smooth horizontal table.
1. Understanding the Given Parameters
Let's list down the information provided in the question:
- Mass of the block (\(m\)) = 20.0 kg
- Horizontal force applied (\(F\)) = 12.0 Newton
- Time duration (\(t\)) = 5.0 seconds
- The table is smooth, which means there is no friction acting on the block.
- The block is placed on the table, implying its initial velocity (\(v_0\)) is 0 m/s.
Our goal is to find the position (which is the displacement, \(x\)) of the block at \(t = 5.0\) seconds.
2. Determining the Acceleration of the Block
To find the position, we first need to determine the acceleration of the block. According to Newton's Second Law of Motion, the force acting on an object is equal to the product of its mass and acceleration. This can be written as:
\($F = ma$\)
Where:
- \(F\) is the force applied (in Newtons)
- \(m\) is the mass of the object (in kilograms)
- \(a\) is the acceleration of the object (in meters per second squared)
We can rearrange this formula to solve for acceleration:
\($a = \frac{F}{m}$\)
Now, let's substitute the given values:
\($a = \frac{12.0 \text{ N}}{20.0 \text{ kg}}$)
\($a = 0.6 \text{ m/s}^2$)
So, the acceleration of the block is 0.6 meters per second squared.
3. Calculating the Block's Position (Displacement)
Since the block starts from rest (\(v_0 = 0\)) and moves with a constant acceleration, we can use the following kinematic equation to find its displacement (\(x\)) after a certain time (\(t\)):
\($x = v_0 t + \frac{1}{2}at^2$\)
Where:
- \(x\) is the displacement (position) of the block (in meters)
- \(v_0\) is the initial velocity (in meters per second)
- \(t\) is the time (in seconds)
- \(a\) is the acceleration (in meters per second squared)
Let's substitute the values we have:
- \(v_0 = 0\) m/s (since it starts from rest)
- \(t = 5.0\) s
- \(a = 0.6\) m/s\(^2\)
Plugging these values into the equation:
\($x = (0 \text{ m/s})(5.0 \text{ s}) + \frac{1}{2}(0.6 \text{ m/s}^2)(5.0 \text{ s})^2$\)
\($x = 0 + \frac{1}{2}(0.6)(25)$\)
\($x = (0.3)(25)$\)
\($x = 7.5 \text{ meters}$\)
Therefore, the position (displacement) of the block at 5.0 seconds is 7.5 meters.
Summary of Steps:
| Step | Description | Formula Used | Calculation |
|---|---|---|---|
| 1 | Identify given values. | N/A | \(m = 20.0\) kg, \(F = 12.0\) N, \(t = 5.0\) s, \(v_0 = 0\) m/s |
| 2 | Calculate acceleration (\(a\)). | \(a = \frac{F}{m}\) | \(a = \frac{12.0}{20.0} = 0.6\) m/s\(^2\) |
| 3 | Calculate displacement (\(x\)). | \(x = v_0 t + \frac{1}{2}at^2\) | \(x = (0)(5.0) + \frac{1}{2}(0.6)(5.0)^2 = 0.3 \times 25 = 7.5\) m |
The calculated position of the block is 7.5 meters.
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