CameraIcon
CameraIcon
SearchIcon
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 ?


Your Answer
A
m=2.08 kg, μs0.402
Your Answer
B
m=1.079 kg, μs0.302
Your Answer
C
m=2.08 kg, μs0.402
Correct Answer
D
m=1.079 kg, μs0.302

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 Upvotes


Similar questions
View More


People also searched for
View More



footer-image