CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

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

A
μMgdcosθcosθ+μsinθ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
μMgdsinθcosθ+μsinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Mgdcosθcosθ+μsinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Mgdsinθcosθ+μsinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A μMgdcosθcosθ+μsinθ
The forces acting on the block are shown in the figure,
As the block moves with uniform velocity the resultant force on it is zero.

Fcosθ=μN .........(1)

Applying equation of equilibrium in vertical direction

Fsinθ+N=Mg ........(2)

Eliminating N from equations (1) and (2),

Fcosθ=μ(MgFsinθ)

Fcosθ+μFsinθ=μMg

F=μMgcosθ+μsinθ

Work done by this force during a horizontal displacement d will be

W=(Fcosθ)d=μMgdcosθcosθ+μsinθ

Therefore, option (a) is correct.

flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image