A body of mass 10 kg moving with a velocity of 1 m/s is acted upon by a force of 50 N for two seconds. The final velocity will be:
11 m/sec
Calculating Final Velocity with Force and Time
This problem asks us to find the final velocity of a body when a constant force acts on it for a specific duration. We are given the initial mass, initial velocity, the magnitude of the applied force, and the time duration the force acts.
To solve this, we can use the principles of dynamics, specifically Newton's Second Law of Motion and the equations of uniformly accelerated motion.
Understanding the Given Information
Let's list the values provided in the question:
- Mass of the body, \(m = 10 \, \text{kg}\)
- Initial velocity of the body, \(u = 1 \, \text{m/s}\)
- Force acting on the body, \(F = 50 \, \text{N}\)
- Time duration for which the force acts, \(t = 2 \, \text{seconds}\)
Calculating Acceleration
When a force acts on a body, it causes the body to accelerate. Newton's Second Law of Motion gives the relationship between force, mass, and acceleration:
\[ F = ma \]
Where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration. We can rearrange this formula to find the acceleration:
\[ a = \frac{F}{m} \]
Let's plug in the given values:
\[ a = \frac{50 \, \text{N}}{10 \, \text{kg}} \]
\[ a = 5 \, \text{m/s}^2 \]
So, the acceleration of the body due to the applied force is \(5 \, \text{m/s}^2\).
Calculating Final Velocity
Now that we have the acceleration, initial velocity, and time, we can find the final velocity using one of the standard equations of motion for constant acceleration. The relevant equation here is:
\[ v = u + at \]
Where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.
Let's substitute the known values into this equation:
\[ v = (1 \, \text{m/s}) + (5 \, \text{m/s}^2)(2 \, \text{s}) \]
First, calculate the product of acceleration and time:
\[ at = (5 \, \text{m/s}^2)(2 \, \text{s}) = 10 \, \text{m/s} \]
Now, add this to the initial velocity:
\[ v = 1 \, \text{m/s} + 10 \, \text{m/s} \]
\[ v = 11 \, \text{m/s} \]
The final velocity of the body after 2 seconds is \(11 \, \text{m/s}\).
Summary of Calculation
| Quantity | Symbol | Value | Formula Used |
|---|---|---|---|
| Mass | \(m\) | \(10 \, \text{kg}\) | Given |
| Initial Velocity | \(u\) | \(1 \, \text{m/s}\) | Given |
| Force | \(F\) | \(50 \, \text{N}\) | Given |
| Time | \(t\) | \(2 \, \text{s}\) | Given |
| Acceleration | \(a\) | \(5 \, \text{m/s}^2\) | \(a = F/m\) |
| Final Velocity | \(v\) | \(11 \, \text{m/s}\) | \(v = u + at\) |
The calculation shows that applying a force of 50 N for 2 seconds to a 10 kg mass initially moving at 1 m/s increases its velocity by 10 m/s, resulting in a final velocity of 11 m/s.
Important Questions from Kinematics and Kinetics
Consider the motion of a point on a circular trajectory. The acceleration in a linear motion (a) and the acceleration in angular motion (α), are related as : (Take r as the radius of circular trajectory)
A ball is dropped on a smooth horizontal surface from height ‘h’. What will be the height of rebounce after second impact, if coefficient of restitution between ball and surface is ‘e’?
Each of four particles move along an x-axis. Their coordinates (in meters) as functions of time (in seconds) are given by
1) particle 1: x (t) = 3.5 – 2.7 t3
2) particle 2: x (t) = 3.5 + 2.7 t3
3) particle 3: x (t) = 3.5 – 2.7 t2
4) particle 4: x (t) = 3.5 – 3.4t - 2.7 t2
Which of these particles have constant acceleration?
If water in a stream is flowing with a velocity of 20 kmph and a boat is travelling from one bank to another bank, if the velocity of boat in a direction perpendicular to direction of stream is 20 kmph and width of the stream is 2km, then the time taken and the angle at which boat makes with the direction stream is,
If a train of length 1000 m is cruising at a speed of 90 kmph and crosses a bridge of length 1000 m, what time does it take to completely pass the bridge?
