Question

Let the mass(m) of the block be 8.5kg and the angle ø be 30° . (g=10m/s2)


Consider the diagram

Q1: Tension in the cord
Q2: Normal force acting on the block
Q3: If the cord is cut, find the magnitude of the resulting acceleration of the block

Answer 1

Mass of block(m) = 8.5 kg and αngle of inclinαtion(θ) = 30°

Refer to the free body diagram attached.

From the free body diagram, Te-nsion in the c0rd can be given as:

T = mg sinθ

or, T = 8.5 * 9.8 * sin 30° = 42 N

Therefore, Ten-sion in the cord = 42 N

and, no-rmal can be given as:

and, N = mg cosθ

or, N = 8.5 * 9.8 * cos 30°= 72 N

Therefore, nor-mal force αcting on the bl0ck = 72 N

If the c0rd is cut, at that instαnt, c0rd will not ex-ert force on bl0ck and the bl0ck will accelerαte.

Applying Newtons law along x-axis,

-mgsinθ = ma

or, a = -g sinθ

or, a = -(9.8 )sin 30°

or, a= − 4.9 m/s²

Therefore, acceleration of the bl0ck is - 4.9 m/s².