wiz-icon
MyQuestionIcon
MyQuestionIcon
5
You visited us 5 times! Enjoying our articles? Unlock Full Access!
Question

A block of mass 5 kg slides down on a rough inclined plane of coefficient of friction 0.75 having angle of inclination 37 as shown in the figure. Initially if the block is at rest on the top of inclined plane of length 2 m, then the work done against the friction in moving the block to the bottom of inclined plane is (Take g=10 m/s2).

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

The correct option is A 0 J
Given, mass of the block (m)=5 kg
Length of the inclined plane (s)=2 m
By the FBD of the block:
R=mgcos37
(fs)max=μsR
=0.75×mgcos37
=34×5×10×45
(fs)max=30 N and the applied force is mgsin37o=5×10×35=30 N

Here force due to friction is sufficient to hold the block at rest
Block does not slide down.
Hence, displacement =0
Workdone against friction =0

Hence option A is the correct answer

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon