Question

An engine of mass $5\times {10}^{4}kg$ pulls a coach of mass $4\times {10}^{4}kg$. A resistance of $1N$ per $100kg$ is acting on both the coach and the engine and if the driving force of the engine is $4500N$ then the acceleration of the engine and the tension in the coupling will respectively be_________

• A. The acceleration of the system (coach$+$engine) is $0.04{m/s}^2$
• B. The acceleration of the system (coach$+$engine) is $0.05{m/s}^2$
• C. The tension in the coupling is $2000N$.
• D. The tension in the coupling is $2400N$.

Answer 1

Answer: (A), (C)

The correct options are
A The acceleration of the system (coach$+$engine) is $0.04{m/s}^2$
C The tension in the coupling is $2000N$.

$Drivingforce=4500N$

Opposing force Resistance $f$

$f=\frac{\left(5+4\right)\times {10}^4}{100}$

$f=900N$

Resultant force:

$F_R=4500-900$

$\Rightarrow F_R=3600N$

Mass of engine and coach $=9\times {10}^{4}kg$

According to Newton's law

$F=ma$

$\therefore 3600=9\times {10}^4\times a$

$\Rightarrow a=\frac{3600}{9\times {10}^4}$

$\Rightarrow a=0.04m/s^2$

Now considering the eqilibrium of the coach only

We have,

$\left(T-R\right)=4\times {10}^4\times 0.04\left(F=ma\right)$

$\Rightarrow T-\frac{4\times {10}^4}{100}=4\times {10}^4\times 0.04$

$\Rightarrow T=(4\times {10}^4\times 0.04)+(4\times {10}^2)$

$\Rightarrow T=1600+400$

$\Rightarrow T=2000N$

Hence, Option A and C are answer.