Questions
A stone weighing 1 kg and sliding on ice with a velocity of 4 m/s is stopped by friction in 30 sec. The force of friction (assuming it to be constant) will be
detailed solution
Correct option is A
u=2 m/s, v=0, t=10 seca=v−ut=0−430=−430=−215=−0.13 m/s2∴Fraction force=ma=1×(−0.13)=−0.13 NTalk to our academic expert!
Similar Questions
Statement I : The familiar equation mg = R for a body on a table is true only if the body is in equilibrium.
Statement II : The equality of mg and R has no connection with the third law.
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
