A boat of mass 700 ,kg is travelling at a spe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given: Mass of the boat, $m = 700 \,kg$. Initial speed, $v_1 = 24 \,ms^{-1}$. Final speed, $v_2 = 6 \,ms^{-1}$. Frictional force, $f = 35v$.Accord

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) Given: Mass of the boat, $m = 700 \,kg$. Initial speed, $v_1 = 24 \,ms^{-1}$. Final speed, $v_2 = 6 \,ms^{-1}$. Frictional force, $f = 35v$.According to Newton's second law, the retarding force is $f = -m \frac{dv}{dt}$.Substituting the given values: $35v = -700 \frac{dv}{dt}$.Rearranging the terms: $\frac{dv}{v} = -\frac{35}{700} dt = -\frac{1}{20} dt$.Integrating both sides from $v_1$ to $v_2$ and $0$ to $t$: $\int_{24}^{6} \frac{dv}{v} = -\int_{0}^{t} \frac{1}{20} dt$.$\ln(\frac{6}{24}) = -\frac{t}{20}$.$\ln(\frac{1}{4}) = -\frac{t}{20}$.$-\ln(4) = -\frac{t}{20} \Rightarrow t = 20 \ln(4) = 20 \ln(2^2) = 40 \ln(2)$.Using $\ln(2) \approx 0.693$, $t = 40 	imes 0.693 = 27.72 \,s$.Rounding to the nearest integer, $t \approx 28 \,s$.

Similar Questions

"When a man walks on a rough surface,the frictional force acts in the direction opposite to his motion." Is this statement correct? Give reasons.

$A$ block of mass $2 \ kg$ is placed on a rough horizontal surface. If a horizontal force of $20 \ N$ acting on the block produces an acceleration of $7 \ m \ s^{-2}$ in it,then the coefficient of kinetic friction between the block and the surface is $(g = 10 \ m \ s^{-2})$.

$A$ body of mass $2 \, kg$ is kept by pressing it against a vertical wall by a force of $100 \, N$. The coefficient of friction between the wall and the body is $0.3$. Then the frictional force is equal to ........ $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module