0

Guest

Courses New

A block of mass 100gm connected to a spring of spring constant 0.1pie^2 N/m is lying on a frictionless horizontal surface. when the compression in the spring is maximum, the block experiences an acceleration of 4pie^2 cm/s^2 away from the spring. If the block moves 2cm from maximum accelerated position in t second, then find the value of ‘t’ .

A block of mass 100gm connected to a spring of spring constant 0.1pie^2 N/m is lying on a frictionless horizontal surface. when the compression in the spring is maximum, the block experiences an acceleration of 4pie^2 cm/s^2 away from the spring. If the block moves 2cm from maximum accelerated position in t second, then find the value of ‘t’ .

Grade:12th pass

1 Answers

venkat
105 Points
6 years ago
Given,
Mass of the block=100gm
Acceleration=4(pie)2
Spring constant k=0.1(pie)2
Initial velocity u=0
When the block and spring system are in equilibrium,
Force due to spring=force on the block
ma=kx
(100gm)*(4(pie)2cm/s2)=(0.1*(pie2)N/m)*(x)
(100*10-3kg)*(4cm/s2)=(0.1N/m)*(x)
(10-1kg)*(4cm/s2)=(10-1N/m)*(x)
x=4cm
Also,
Displacement s=2cm
s=ut+1/2(at2)
2cm=0+1/2(4(pie)2)t2
t2 =1/(pie)2
t=1/(pie)
 

Think You Can Provide A Better Answer ?

Other Related Questions on Engineering Entrance Exams

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More