logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

Calculate the magnitude of the normal force and the acceleration of the box in this experiment: Take a 10 kg box on a frictionless 30o inclined plane and release your hold, allowing the box to slide to the ground, a horizontal distance of d meters and a vertical distance of h meters as shown in the diagram below:

$\begin{array}{1 1}45N,9.8m/s^2\\85N,9.8m/s^2\\45N,4.9m/s^2\\85N,4.9m/s^2\end{array} $

1 Answer

Answer : $85N,4.9m/s^2$
Free body diagram :
The normal force must be equal to the y-component of the gravitational force (weight). Calculating the normal force is then just a matter of finding the y-component of the gravitational force = $mg \cos 30^{\large\circ} = 85$ N
We know that the force pulling the box in the positive x direction has a magnitude of $mg \sin 30^{\large\circ}$ – which is the x-component of the weight. Using Newton’s Second Law, F = ma, we just need to solve for a which is = $g \sin\theta = 9.8 \times \sin 30^{\large\circ} = 4.9 m/s^2$
answered Aug 27, 2014 by sreemathi.v
 

Related questions

...