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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Vector Algebra

A tow truck drags a stationary car along the road. The chain makes an angle of 30 degrees with the road and the tension in the chain is 1500 N. How much work is done (in Joules) by the truck in pulling the car 1 km (approximate answer, rounded off)?

(A) 1299038 J

(B) 1400004 J

(C) 2300032 J

(D) 12344548 J

1 Answer

Toolbox:
  • $\overrightarrow a.\overrightarrow b=|\overrightarrow a|\:|\overrightarrow b|\:cos\theta$
Given the Force, $\overrightarrow F$ and Distance $\overrightarrow D$ vectors, we can write the work done as $\overrightarrow W = \overrightarrow F \times \overrightarrow D$
$\overrightarrow a.\overrightarrow b=|\overrightarrow a|\:|\overrightarrow b|\:cos\theta$
$\Rightarrow$$\;\overrightarrow W = \overrightarrow F \times \overrightarrow D = | \overrightarrow F| \times |\overrightarrow D| \times \cos \theta$
In our case $\theta = 30 ^{0} = \large\frac{\pi}{6}$
$\Rightarrow$$\;\overrightarrow W = | \overrightarrow F| \times |\overrightarrow D| \times \cos \theta = 1500N \times 1000m \times \cos \large(\frac{\pi}{6})$
$\Rightarrow \overrightarrow W = 750000 \sqrt 3 J \approx 1299038 J$
answered Feb 11, 2014 by balaji.thirumalai
 

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