wiz-icon
MyQuestionIcon
MyQuestionIcon
17
You visited us 17 times! Enjoying our articles? Unlock Full Access!
Question

A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with the horizontal. The friction coefficient between the block and the surface is μ. If the block travels at a uniform velocity, then the work done by this applied force during a displacement d of the block is

A
W=μMg dcosθcosθ+μsinθ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
W=μMg dcosθsinθ+μcosθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
W=μMg dsinθsinθ+μcosθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
W=μMg dsinθsinθ+μcosθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A W=μMg dcosθcosθ+μsinθ


Resolving forces in vertical direction

N=MgFsinθ(1)
Resolving forces in horizontal direction:
Fcosθ=fk=μN(2)
From eq. (1) and (2)
Fcosθ=μ[MgFsinθ]
Fcosθ+μFsinθ=μMg
F=μMgcosθ+μsinθ
As work done W=Fdcosθ
substituting the value of F in above equation we get,
W=μMg dcosθcosθ+μsinθ

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
So You Think You Know Work?
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon