Question

A man in a boat a pulls a rope with a force 100n. The other end of the rope is tied to a boat b of mass 200 kg. The total mass of boat a and man is 300kg, disregard the weight of the rope and the resistance of the water. The power developed by the man by the end of the third second is

Answer 1

Answer: 250 N

Explanation:

Power = Force X Relative Velocity

p = F X $V_{rel}$

Force = mass X acceleration

In order to find acceleration of the boat

$F_b$ = $m_b$ . $a_b$

100 = 200 . $a_b$

$a_b$ = $\frac{-1}{2}$  $\frac{m}{s^2}$

Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v.

$v_b$ = $u_b$ + $a_b$.T

$v_b$ = $\frac{1}{2}$ X 3 (Since, Initial velocity is zero)

$v_b$ = $\frac{3}{2}$ $\frac{m}{s}$   .............equation 1

so,

$v_{rel}$ + $v_{a}$ = $v_b$

$m_bv_b$ + $m_av_a$ = 0

$m_bv_b$  + $m_a$($v_{b} - v_{rel}$) = 0

200$v_b$ + 300($v_{b} - v_{rel}$) = 0

500$v_b$ = 300$v_{rel}$

Substituting value of $v_b$ from equation 1 , we get

$v_{rel}$ = 2.5 $\frac{m}{s}$

∴P = f X $v_{rel}$

= 100 X 2.5

=250 N

∴Power generated at the end of third second is 250 N

Answer 2

Thus the power developed by the man is P = 250 N

Explanation:

Given data:

• Force "f" = 100 N
• Mass of boat = 200 Kg
• Total mass = 300 Kg

Solution:

P = F . V(red)

FB = mBaB

100 = 200 aB

aB = 1/2 m /s^2

VB = UB + aBt

VB = 1/2 x 3

VB = 3/2 m/s^2

V(rel) + VA = VB

mBvB + mAvA = 0

mBvB + mA (vB - vrel) = 0

200 vB + 300 (vB - vrel) = 0

500 vB = 300 Vrel

500 x 3/2 = 300 Vrel

Vrel = 5 / 2 = 2.5 m/s

P = 100 x 2.5 = 250 N

Thus the power developed by the man is P = 250 N