# Thread: mechanics- block on wedge 2

1. ## mechanics- block on wedge 2

Hi, I recently attempted this mechanics past exam question but think I have done it wrong because the questions are all worth 5 marks and my answers seem far too short (particularly parts b and c) though I cannot see at all why they would be wrong, can anyone see what am I doing wrong?

A wedge with inclination a rests next to a wall. A block of mass m slides down the plane. There is no friction between the block and the wedge or the wedge and the horizontal surface. The block is initially at rest at the top of the wedge and the height of the wedge is h. (each part is worth 5 marks).

a). Explain why the net force on the block points down the incline?
Only forces on block are weight and normal force from the wedge.

Component of weight down slope mgsina (this is unapposed)

component of weight perpendicular to slope mgcosa this is equal and opposite to normal. Hence net force =mgsina down slope.

b).Find an expression for the sum of all the forces acting on the block.
I thought this was just mgsina

c.) Find the normal force exerted by the wedge on the block

d). Show that the magnitude of the normal force exerted by the wall on the wedge is mgcosasina.
My answer: normal force (on block) =mgcosa (from c)
Horizontal component of this is -mgcosasina and the normal force from the wall on the wedge must be equal and opposite.

e). What will be the speed of the block when it reaches the bottom of the wedge.

mgh=0.5mv^2 Therefore v=sqrt(2gh).

2. First of all Did you draw the free body diagram showing all the forces correctly, it will be considered most important

3. I didn't know I needed to. Would that be for part a? I have now drawn it but don't have a scanner so I cannot display it in the thread though I think it agrees with my previous answers.