$A$ constant force acts on an object of $5 \ kg$ for a period of $2 \ s$. It increases the velocity of the object from $3 \ m s^{-1}$ to $7 \ m s^{-1}$. Find the magnitude of the applied force. Now,if the force were applied for a period of $5 \ s$,what would be the final velocity of the object?
(D) Given: Mass $m = 5 \ kg$,initial velocity $u = 3 \ m s^{-1}$,final velocity $v = 7 \ m s^{-1}$,time $t = 2 \ s$.
Using Newton's second law of motion,$F = m \times a = m \times \frac{(v - u)}{t}$.
Substituting the values: $F = \frac{5 \times (7 - 3)}{2} = \frac{5 \times 4}{2} = 10 \ N$.
Now,for the second part,the force $F = 10 \ N$ is applied for $t' = 5 \ s$.
The acceleration $a = \frac{F}{m} = \frac{10}{5} = 2 \ m s^{-2}$.
Using the first equation of motion,$v' = u + a \times t'$.
$v' = 3 + (2 \times 5) = 3 + 10 = 13 \ m s^{-1}$.
Using Newton's second law of motion,$F = m \times a = m \times \frac{(v - u)}{t}$.
Substituting the values: $F = \frac{5 \times (7 - 3)}{2} = \frac{5 \times 4}{2} = 10 \ N$.
Now,for the second part,the force $F = 10 \ N$ is applied for $t' = 5 \ s$.
The acceleration $a = \frac{F}{m} = \frac{10}{5} = 2 \ m s^{-2}$.
Using the first equation of motion,$v' = u + a \times t'$.
$v' = 3 + (2 \times 5) = 3 + 10 = 13 \ m s^{-1}$.
Explore More
Similar Questions
Give reasons and state the laws related to these statements:
$(a)$ It is easier to push an empty box than to push a box full of books.
$(b)$ It is difficult for a fireman to hold a hose which ejects a large amount of water with high velocity.
$(a)$ It is easier to push an empty box than to push a box full of books.
$(b)$ It is difficult for a fireman to hold a hose which ejects a large amount of water with high velocity.
Medium
View SolutionWhat force would be needed to produce an acceleration of $4 \, m s^{-2}$ on a ball of mass $6 \, kg$?
Easy
View SolutionState whether the following statement is true or false:
$1 \ N$ is the force that produces an acceleration of $1 \ m \ s^{-2}$ in a body of mass $1 \ g$.
$1 \ N$ is the force that produces an acceleration of $1 \ m \ s^{-2}$ in a body of mass $1 \ g$.
Easy
View Solution$A$ man throws a ball of mass $0.5\, kg$ vertically upward with a velocity of $25\, m s^{-1}$. Find:
$(a)$ The initial momentum of the ball.
$(b)$ The momentum of the ball at the halfway mark of the maximum height (given $g = 10\, m s^{-2}$).
$(a)$ The initial momentum of the ball.
$(b)$ The momentum of the ball at the halfway mark of the maximum height (given $g = 10\, m s^{-2}$).
Difficult
View Solution$(a)$ Using Newton's law of motion, derive the relation between force and acceleration.
$(b)$ Define one newton.
$(c)$ Which would require a greater force to accelerate: a $0.5 \, kg$ mass at $5 \, m s^{-2}$ or a $4 \, kg$ mass at $2 \, m s^{-2}$? Give reason.
$(b)$ Define one newton.
$(c)$ Which would require a greater force to accelerate: a $0.5 \, kg$ mass at $5 \, m s^{-2}$ or a $4 \, kg$ mass at $2 \, m s^{-2}$? Give reason.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo