Question

A block of mass 2 kg slides down the face of a smooth 45 degrees wedge of 9 kg the wedge is placed at a frictionless horizontal surface

Answer 1

Answer:

Acceleration = 1 m/s^2

Explanation:

There are three forces acting on the block. One is acting on the base of the wedge, other on the opposite direction and the third is along the direction of block movement.

Let m be the mass of block, M mass of wedge and the angle would be 45 .

Equation for resultant of forces acting is;

Mg cos 45 – N = mA cos 45…………..eq(1)

First we calculate N;

N cos 45 = M(A)

N = M(A)/cos 45

Putting value of N is eq (1)

Mg cos 45 – M(A)/cos45 = mA cos 45

Putting values of m and M and g;

2 x 10 x 1/(√2)  - 9 x A x √2  = 2 A 1/(√2)  

Simplyfiying above;

10 √2 = 10 √2 x A

Hence;

A = 1 m/s^2