Question

A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the particle between t=0.00 seconds and t=5.00 seconds. find the speed v of the particle at t=5.00 seconds. express your answer in meters per second to three significant figures.

Answer 1

The speed of the particle at $t = 5\,{\text{s}}$ is $\fbox{43\,{\text{m/s}}}$.

Further explanation:

Applied force is directly proportional to the mass of the body and the acceleration of the body.

Given:

The function for force is $F\left( t \right) = 6{t^2} - 4t + 3$.

The limit of time is $t = 0\,{\text{s}}$ to $t = 5\,{\text{s}}$.

The mass of the particle is $5\,{\text{Kg}}$.

Concept used:

The rate of change of displacement of a body in unit time is called speed of the body. It is a scalar quantity.

Newton’s second law of motion states that the rate of change of momentum is equal to the applied force.

The expression for the Newton’s second law is given as.

$F=ma$

Rearrange the expression for the acceleration of the body.

$a=\dfrac{F}{m}$                                                              …… (1)

Here, a is the acceleration of the body, F is the force applied and m is the mass of the body.

The rate of change of speed of a body in unit time is called acceleration of the body.  

The expression for the acceleration is given as.

$a=\dfrac{{dv}}{{dt}}$

 

Substitute $\dfrac{{dv}}{{dt}}$ for $a$ in equation (1).

$\dfrac{{dv}}{{dt}} = \dfrac{F}{m}$

 

On integrating the above expression we get.

$v = \dfrac{1}{m}\int\limits_0^5 {Fdt}$                                 …… (2)

Substitute $6{t^2} - 4t + 3$ for$F\left(t\right)$ and $5\,{\text{Kg}}$ for m in equation (2).

$\begin{aligned} v&=\frac{1}{{\left( {5\,{\text{Kg}}} \right)}}\int\limits_0^5 {\left( {6{t^2} - 4t + 3} \right)dt}\\&=\frac{1}{{\left( {5\,{\text{Kg}}}\right)}}\left|{3{t^3}-2{t^2}+ 3t}\right|_0^5\\&=\frac{{215}}{{5\,}}\,{\text{m/s}}\\&=43\,{\text{m/s}}\\\end{aligned}$

 

Thus, the speed of the particle is $\fbox{43\,{\text{m/s}}}$.

Learn more:

1.  Motion under force brainly.com/question/4033012.

2.  Conservation of momentum brainly.com/question/9484203.

3. Motion under friction brainly.com/question/7031524.

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Acceleration, force, weight, mass, motion, impact, nail, hammer, acceleration, duration of impact, Newton’s laws, speed, rate of change, time , 43m/s, 5Kg.

Answer 2

v =Integral(F/m)dtv=Integral(F/m)dt =â«50(6t2â’4t+3)/5dt=â«05(6t2â’4t+3)/5dt =1/5â—(2t3â’2t2+3t)|(0,5)=1/5â—(2t3â’2t2+3t)|(0,5) =1/5â—(250â’50+15)=1/5â—(250â’50+15) =215/5=215/5 =43m/s.