Solveeit Logo

Question

Question: A \[15{\text{ }}kg\] mass is accelerated from rest with a force of \[100N\]. As it moves faster, fri...

A 15 kg15{\text{ }}kg mass is accelerated from rest with a force of 100N100N. As it moves faster, friction and air resistance create an oppositely directed retarding force given by FR=A+BV{F_R} = A + BV , where A=25 NA = 25{\text{ }}N and B=0.5Nm/sB = 0.5\dfrac{N}{{m/s}} . At what velocity does the acceleration equal to one half of the initial acceleration?
A. 25 m/s{\text{A}}{\text{. }}25{\text{ }}m/s
B. 50 m/s{\text{B}}{\text{. }}50{\text{ }}m/s
C. 75 m/s{\text{C}}{\text{. }}75{\text{ }}m/s
D. 100 m/s{\text{D}}{\text{. }}100{\text{ }}m/s

Explanation

Solution

Calculate the initial acceleration from the given mass and the force.
Find the relation between the velocity and the force by putting the given values of the constant term in the given equation.
Since we need to calculate the velocity for the acceleration would be one and half of the initial acceleration, write the force in terms of acceleration in the equation, and find the value of the velocity.

Formula used:
F=m×aoF = m \times {a_o}
a0=Fm\Rightarrow {a_0} = \dfrac{F}{m}
Where a0{a_0} = the initial acceleration.
mm= the mass of the object.
F=F = the force when the mass was at rest.
a=a0+FRma = {a_0} + \dfrac{{{F_R}}}{m}
Where,
aa = the acceleration at a certain velocity when the object starts moving.
FR{F_R} = the retarding force

Complete step by step answer:
Newton’s 2nd{2^{nd}} law of motion defines the force as the product of the mass and acceleration of an object. That means if a force FF is applied on an object of mass mm moving with acceleration aa, the equation will be, F=m×aF = m \times a.
In the given problem, when the object is at rest the force is applied on an object,
so F=m×aoF = m \times {a_o}
a0=Fm\Rightarrow {a_0} = \dfrac{F}{m}
Where a0{a_0} = the initial acceleration.
mm= the mass of the object = 15 kg15{\text{ }}kg
F=F = the force when the mass was at rest = 100N100N
a0=10015m/s2\therefore {a_0} = \dfrac{{100}}{{15}}m/{s^2}
Now, when the object starts moving the friction and the air resistance create opposite retarding force given by,
FR=A+BV{F_R} = A + BV………………(11)
where V=V = velocity at any instant.
A=25 NA = 25{\text{ }}N
B=0.5Nm/sB = 0.5\dfrac{N}{{m/s}}
Putting the value of AA and BB in the equation (11),
FR=25+0.5V\therefore {F_R} = 25 + 0.5V……………(22)
Now, the acceleration at a certain velocity when the object starts moving,
a=a0+FRma = {a_0} + \dfrac{{{F_R}}}{m}…………………….(33)
Given, a=3a02a = \dfrac{{3{a_0}}}{2} [since 1+12=321 + \dfrac{1}{2} = \dfrac{3}{2} ]
Put the value of aa we get from eq. (33)
3a02=a0+FRm\Rightarrow \dfrac{{3{a_0}}}{2} = {a_0} + \dfrac{{{F_R}}}{m}
FRm=a02\Rightarrow \dfrac{{{F_R}}}{m} = \dfrac{{{a_0}}}{2}
FR=ma02\Rightarrow {F_R} = \dfrac{{m{a_0}}}{2}
FR=15×1002×15\Rightarrow {F_R} = \dfrac{{15 \times 100}}{{2 \times 15}}[since mm = 15 kg15{\text{ }}kgand a0=10015m/s2{a_0} = \dfrac{{100}}{{15}}m/{s^2}]
on dividing the term and we get,
FR=50\Rightarrow {F_R} = 50
Putting the value of FR{F_R} in the equation (22),
50=25+0.5V\Rightarrow 50 = 25 + 0.5V
0.5V=5025\Rightarrow 0.5V = 50 - 25
V=250.5\Rightarrow V = \dfrac{{25}}{{0.5}}
V=50m/s\Rightarrow V = 50m/s
So, when the velocity is V=50m/sV = 50m/s, the acceleration will be one and half of the initial acceleration.

Hence, the right answer is in option   (B)\;\left( B \right).

Note: Here we take the acceleration at a certain velocity which is the sum of the initial acceleration i.e when the mass is at rest and the acceleration due to the retarding force.
The acceleration due to the retarding force is also taken based on Newton’s  2nd\;{2^{nd}} law i.e acceleration is the applied force per unit mass.