CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

With the assumption of no slipping determine the mass m of the block which must be placed on the top of a 5 kg cart in order that the time period of oscillation is 2 sec. What is the minimum coefficient of static friction μs, for which the block will not slip relative to the cart, if the cart is displaced 300 mm, from the equilibrium position and released ?

A
m=2.08 kg, μs0.402
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
m=1.079 kg, μs0.302
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
m=2.08 kg, μs0.402
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
m=1.079 kg, μs0.302
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D m=1.079 kg, μs0.302
As we know that
T=2π5+m60
2=2π5+m60
m=1.079 kg
For maximum acceleration of SHM.
amax=ω2A
So maximum force an mass 'm' is m amax=mω2A. and this mω2A is provided by friction force only
So μs mg m ω2 A
μsω2Ag
μs(2πT)2Ag
μs(2π2)2×0.39.8
or
μs0.302

flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image