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