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

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.

Step-by-step 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* 9.8* \sin(30)-F\\F_(net)=73.5-F`

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

`F_(net)=ma\\73.5-F=15* 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.