Question

A man in a boat A, the mass of which is m1= 250kg, pulls a rope with a force of F= 200N.
Case1 the other end of the rope is tied to a tree on the shore,
Case2 the other end to a boat B having a mass of 200 kg.
Find in both cases the velocity of boat A at the end of the third second. What is the work done in this time and what is the power developed by the man in both cases at the end of the third second?

Answer 1

Hint: In order to solve the question, we will use the formula of acceleration, work done, force and power .so that we can answer all the subparts asked in the question we will solve both cases together so as make the question simply as we can compare and use the common quantities.

Formula Used:
$a = \dfrac{v}{t}$
V refers to velocity
t refers to time
a refers to acceleration
$F = ma$
F refers to force
m refers to mass
a refers to acceleration
$W = \dfrac{1}{2}m{v^2}$
W refers to work done
m refers to mass
v refers to velocity
$P = Fv$
P refers to power
F refers to force
v refers to velocity

Complete step by step answer:
We will solve both the cases to compare the situation simultaneously for the question we have to find the velocity, work done and the power.
In the question we are given
Boat A of Mass m1 = 250 kg pull a rope with force F=200 N
While boat B is of mass m=150 kg
We are given two cases
Case1 says boat A is pulling a tree
Case2 says boat A is pulling a boat B
In both the cases the acceleration of the boat A will be same that we can find using the formula
$F = ma$
$a = \dfrac{F}{{m_1}}$
Substituting the F and m1
$a = \dfrac{{200}}{{250}}m{s^{ - 2}}$
$a = \dfrac{4}{5}m{s^{ - 2}}$
$a = 0.8{\text{ }}m{s^{ - 2}}$
Hence acceleration of boat A is $0.8{\text{ }}m{s^{ - 2}}$
And the velocity of Boat A will also be same by using the formula
$a = \dfrac{v}{t}$
$v = at$ ……. (1)
$a = \dfrac{F}{{m_1}}$
Substituting value of a in equation 1
$v = \dfrac{{F \times t}}{{m_1}}$
t is given in question as 3 second
$v = \dfrac{{200 \times 3}}{{250}}$
$v = 2.4{\text{ m}}{{\text{s}}^{ - 2}}$
Hence we will name it as ${v_1}$
${v_1} = 2.4{\text{ m}}{{\text{s}}^{ - 2}}$
By using we will find velocity of boat b
${v_2} = \dfrac{{F \times t}}{{m2}}$
t is given in question as 3 second
${v_2} = \dfrac{{200 \times 3}}{{150}}$
${v_2} = 4{\text{ m}}{{\text{s}}^{ - 2}}$
The work done by man in first case
${W_1} = \dfrac{1}{2}{m_1}{(v_1)^2}$
Substituting value of $v_1$ and $m_1$
${W_1} = \dfrac{1}{2}250{(2.4)^2}$
${W_1} = 720{\text{ Nm}}$
The work done by man in second case
${W_2} = \dfrac{1}{2}{m_1}{(v_2)^2} + \dfrac{1}{2}m_2{(v_2)^2}$
Substituting value of $v_2$, $m_2$ and $m_1$
${W_2} = \dfrac{1}{2}250{(4)^2} + \dfrac{1}{2}150{(4)^2}$
${W_2} = {\text{ 3200Nm}}$
Now we will find the power for both the cases using the formula
$P = Fv$
First case
${P_1} = F{v_1}$
Substituting the values, we will get
${P_1} = (200 \times 2.4)W$
${P_1} = 480{\text{ }}W$
Second case
${P_2} = F({v_1} + {v_2})$
Substituting the values, we will get
${P_2} = (200 \times 6.4)W$
${P_2} = 1280{\text{ }}W$
Hence, the answers are
 ${v_1} = 2.4{\text{ m}}{{\text{s}}^{ - 2}}$
${v_2} = 4{\text{ m}}{{\text{s}}^{ - 2}}$
${W_1} = 720{\text{ Nm}}$
${W_2} = {\text{ 3200Nm}}$
${P_1} = 480{\text{ }}W$
${P_1} = 1280{\text{ }}W$

Note: Many of the people will make the by note using the correct formula. It is very important that in work done and power we have to use these formulas to get to the answer along with we have to take care of the unit of measurement so as to avoid any confusion in the answer.