Question

A block of mass 2 kg connect with. A spring of natural length40 cm of k =200. Coff.. of friction is. 0 .5. When released from the given position accelerate of block ..,......



And 6 m/sec2

Answer 1

Answer:

6m/s^2

Explanation:

F=kx

But as this Force is acting at Theta angle with vertical we will use

F=kx sin(Theta)

Put the values

F=1200N

And F=ma

From here

a=600cm/s^2

But it is in cm so converting into m/s^2

=6m/s^2