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

Answer:

The speed will be "16.67 m/s".

Explanation:

The given values are:

Distance

= 72 m

Angle

= 30°

Acceleration

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

                    = $(9.8\times sin30^{\circ}) - (0.53\times cos30^{\circ})$

                    = $1.929 \ m/s^2$

Let the speed be "v".

⇒  $v^2=u^2+2as$

⇒  $v^2=0(2\times 1.929\times 72)$

⇒  $v^2=277.226$

⇒  $v=\sqrt{277.776}$

⇒  $v=16.67 \ m/s$