1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A 30 kg block rests on a rough horizontal surface. A horizontal force of 200 N is applied on the body. The block acquires a speed of 4 m/sec, starting from rest, in 2 sec. What is the value of coefficient of friction?

A
103
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
310
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.47
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0.185
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 0.47 So equation for the motion is, F−fk=ma where fk=μR=μmg=μ×30×9.8=294μ Since initial velocity of the body is zero & its velocity increases from zero to 4 m/sec in 2 seconds, so acceleration of the body is given by, v=u+at ⇒4=0+a×2 or a=2 m/sec2 So, F−fk=30×2 ⇒200−μ×294=60 ⇒μ=140294=0.476

Suggest Corrections
6
Join BYJU'S Learning Program
Related Videos
Friction: A Quantitative Picture
PHYSICS
Watch in App
Join BYJU'S Learning Program