Question

A box is sliding down an incline tilted at a 11.8° angle above horizontal. The box is initially sliding down the incline at a speed of 1.60 m/s. The coefficient of kinetic friction between the box and the incline is 0.390. How far does the box slide down the incline before coming to rest?

A) .736 m

B) 1.03 m

C) .883 m

D) .589 m

E) the box does not stop. it accelerates down the plane.

Answer 1

Sol::

Given data:

Angle ( Ѳ )= 11.8°
speed (Vi) = 1.60 m/s
coefficient of kinetic friction =0.390

The x, y components of the weight of the block are

Fx= mg*sin11.8°
Fy= mgcos11.8°

As the normal force on the block is counter balanced by Y component of weight of block
So
Fn= mg*cos 11.8°

Then the friction force we will be

Ff=u*Fn
= 0.39*mgcos11.8°

So the net force along the inclined plane will be

Fnet= mgsin11.8°- Ff
= mgsin11.8°-0.39*mgcos11.8°

So the acceleration of block is given by using Newton's II law

a= Fnet/m
= mgsin11.8°-0.39*mgcos11.8°/m
=gsin11.8°-0.39*gcos11.8°
= 9.8*sin11.8-0.39*9.8*cos11.8°
= -1.737 m/s²

now by using kinematics we get

Vf²-Vi²= 2ad
0-1.6²= 2(-1.737)*d
d= 0.736m

So option (A) is correct

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