Question

An object with a mass of 5.5 kg is allowed to slide from rest down an inclined plane. The plane makes an angle of 30o with the horizontal and is 72 m long. The coefficient of friction between the plane and the object is 0.35. The speed of the object at the bottom of the plane is:_________.

a. 24 m/s.
b. 11 m/s.
c. 15 m/s.
d. 5.3 m/s.
e. 17 m/s.

Answer 1

The speed will be "16.67 m/s".

Step-by-step explanation:

The given values are:

Distance

= 72 m

Angle

= 30°

Acceleration

=
`g(sin \theta-ucos \theta)`

=
`(9.8* sin30^(\circ)) - (0.53* cos30^(\circ))`

=
`1.929 \ m/s^2`

Let the speed be "v".

⇒
`v^2=u^2+2as`

⇒
`v^2=0(2* 1.929* 72)`

⇒
`v^2=277.226`

⇒
`v=√(277.776)`

⇒
`v=16.67 \ m/s`