A Physics Question
Explain it clearly.
1. A block of mass 5kg is placed on a rough inclined plane. The friction between the block and the inclined plane is 10N.
a) Draw a labelled diagram to show all the forces acting on the block.
b) Find the net force acting on the block, and hence describe the motion of the block.
c) Calculate the
i) loss in p.e. of the block
ii) work done against friction
iii) gain in k.e. of the block
when it reaches the ground.
d) Calculate the speed of the block when it reaches the ground.
e) Explain whether the work done by normal reaction is positive, negative, or zero.
- 1 decade agoFavorite Answer
There are some missing info from the question. what is the inclined angle (A) of the plane? how far is the block from the ground? whether the block can keep the position or not (I assume it cannot keep it's position and begins to move downwards)?
a. all forces acting on block: 1. the weight of the block 50N downward.
2. the reaction from the plane to the block 50cosA N. 3. the friction of 10N alone the incline plane.
b. Net force acting on the block:
50sinA - 10 down the slope.
sine F = ma
a = (50sinA-10)/10
thus the block will move down the slope at an acceleration of a.
c. loss in pe = mgh where h is the vertical distance between the initial position of the block and the ground.
work done against friction = Fs where s is the distance the block had travelled.
KE when it reach ground = PE - Fs, i.e. loss in PE will transform into KE of the block + work done due to friction.
Just fit in the missing figures into the above and you will get the answer.