Physics, asked by naranbhainayak294, 3 months ago

A body of mass 10 kg is moved parallel to the ground (a) 196j (b)- 196j. (c) 10 j. (d) zero

Answers

Answered by trilakshitha
0

Answer:

Given,

Mass

M=

10kg

Work done,

W=

150J

Displcement on horizontal = displacement on incline

=

d=

2m

Friction force on horizontal plane, F

r

= μmg

Friction force on incline plane, F

r

′= μmgcosθ

Speed is uniform means net acting force is equal to friction force

Work done,

W=

F

r

.d=

μmg.d

⇒ μ=

mg.d

W

=

10×10×2

150

= 0.75

Work done on incline plane,

W=

F

r

′d=

μmgcosθ.d

W=

0.75×

10×

10×

cos

30

o

×

2

W=

75

3

J

Work done against friction is

75

3

J

Explanation:

Hi...

Hope it helps you...

Answered by abhinavkumar1950
0

Explanation:

Given,

Mass M=10kg

Work done, W=150J

Displcement on horizontal = displacement on incline =d=2m

Friction force on horizontal plane,F

r

=μmg

Friction force on incline plane,F

r

=μmgcosθ

Speed is uniform means net acting force is equal to friction force

Work done,W=F

r

.d=μmg.d

⇒μ=

mg.d

W

=

10×10×2

150

=0.75

Work done on incline plane, W=F

r

d=μmgcosθ.d

⇒W=0.75×10×10×cos30

o

×2

⇒W=75

3

J

Work done against friction is 75

3

J

HOPE ITS HELP!!

Please mark me as brainlist

Similar questions