Question

A friction is inclined plane makes an angle of 30° with the horizontal find the constant force applied parallel to the plane required to causeif 15 kg box to slide off the plane with an acceleration of 1.2 m/s

Answer 1

Answer:

The constant force applied parallel to the plane required to cause a 15 kg box to slide off the plane with an acceleration of 1.2 m/s is 55.5 N.

Explanation:

Given:

Mass of the box is, $m=15\ kg$

Acceleration along the plane is, $a=1.2\ m/s$

Angle of inclination of plane is, $\theta=30$°

The given inclined plane is friction-less. Let the constant force applied be $F$ N.

Now, consider the free body diagram of the box as shown below.

The forces acting along the plane are the constant force, $F$ up the plane and $mg\sin \theta$ down the plane.

The forces acting perpendicular to the plane are the normal force $N$ and $mg\cos \theta$.

Therefore, net force along the direction of the inclined plane is given as:

$F_{net}=mg\sin \theta-F\\F_{net}=15\times 9.8\times \sin(30)-F\\F_{net}=73.5-F$

Now, as per Newton's second law of motion,

$F_{net}=ma\\73.5-F=15\times 1.2\\73.5-F=18\\F=73.5-18=55.5\ N$

Therefore, the constant force applied parallel to the plane required to cause a 15 kg box to slide off the plane with an acceleration of 1.2 m/s is 55.5 N.