Question

A body of mass 0.3 kg is taken up an inclined plane of length 10 m and height 5 m, and then allowed to slide down the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the
(i) work done by the applied force over the upward journey?
(ii) work done by gravitational force over the round trip?
(iii) work done by the frictional force over the round trip?

Which of the above forces (except applied force) is/are
conservative forces?

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Answer 1

Given :-

Mass of the body, m = 0.3 kg

Length of the plane, l = 10 m

Height of the plane, h = 5 m

Coefficient of friction, μ = 0.15

Relations :-

• N = mgcosθ

• F = mgsinθ + f [f = μN = μmgcosθ]

• F = mgsinθ + μmgcosθ

• As, length is 10 m, the value of sinθ = $\dfrac{5}{10}$ = $\dfrac{1}{2}$ and the value of cosθ = $\dfrac{√3}{2}$ as, sin 30° = $\dfrac{1}{2}$ and cos 30° = $\dfrac{√3}{2}$

Solution :-

i. Work done by the applied force over the upward journey, $W_{up}$ = F×d

• $W_{up}$ = (mgsinθ + μmgcosθ)10

• $W_{up}$ = 10mg(sinθ + μcosθ)

• $W_{up}$ = 10×0.3×9.8 [½+ 0.15($\dfrac{√3}{2}$)]

• $W_{up}$ = 29.4 [ 0.5 + 0.129 ]

• $W_{up}$ = 29.4 [ 0.62 ]

• $W_{up}$ = 18.5 J

ii. Work done by gravitational force over the round trip, $W_{g}$ = -mg∆h

• $W_{g}$ = -(0.3)(9.8)(0)

• $W_{g}$ = 0

The displacement over the round trip is zero, the object ended it's journey from where it started so, work done by the gravitational force over the round trip is zero.

iii. Work done by the frictional force over the round trip, $W_{f}$ = fscos(180°)

• $W_{f}$ = (μmgcosθ)(20)(-1)

• $W_{f}$ = (0.15)(0.3)(9.8)($\dfrac{√3}{2}$)(20)(-1)

• $W_{f}$ = -(0.441)(0.86)(20)

• $W_{f}$ = -(0.38)(20)

• $W_{f}$ = -7.6 J

Gravitational force is conservative force, as in the solution ii. I mentioned about change in height, which states that it is not path dependent.

$\bold{Hope\;it \; helps\;!}$

Answer 2

Given:

• Mass of the body (m) = 0.3 kg
• Length of the inclined plane (l) = 10 m
• Height of the plane (h) = 5 m
• Coefficient of kinetic friction (μₖ)  = 0.15

Solution:

First, we need to find the angle of inclination,

sin θ  = h / l = 5/10 = 0.5

θ  = 30

(i) work done by the applied force over the upward journey?

⇒ F  = mg sin θ  +  fₖ

⇒ F  = mg sin θ  +  μₖ N

⇒ F = mg sin θ  +  μₖ mg cos θ

⇒ F =  mg ( sin θ +  μₖ  cos θ)

⇒ F =  0.3 x 9.8 ( sin 30 +  0.15x    cos 30)

⇒ F = 2.94 ( 0.5 + 0.15 x 0.86)

⇒ F= 2.94( 0.5 + 0.129)

⇒ F = 2.94 ( 0.629)

⇒ F = 1.84 N

⇒ Work done = F  x l

⇒ Work done = 1.84 x 10

⇒ Work done = 18.4 J

(ii) work done by gravitational force over the round trip?

Work done by the gravitational force when the block is moving in the upward direction (against gravity)

⇒ W = $\sf{F_g}$  h

⇒ W =  - m g sin θ x h

⇒ W = - 0.3 x 9.8 x 0.5 x 5

⇒ W = - 7.35 J

Work done by the gravitational force when the block is moving in the downward direction( along gravity)

⇒ W' = $\sf{F_g}$' h

⇒ W' = mg sin  θ x h

⇒ W' = 0.3 x 9.8 x 0.5 x 5

⇒ W' = + 7.35 J

Net work done = W + W' = -7.35 + 7.35 = 0

Hence, work done by gravitational force over the round trip is 0

(iii) work done by the frictional force over the round trip?

Frictional force always opposes the direction of motion hence it will always be negative

Work done by the frictional force when the block is moving in upward direction

⇒ W = -  fₖ .l

⇒ W = - μₖ mg cos θ x l

⇒ W = - 0.15 x 0.3 x 9.8 x 0.86 x 10

⇒ W = - 3.79 J

Work done by the frictional force when the block is moving in downward direction

⇒ W' = -  fₖ .l

⇒ W' = - μₖ mg cos θ x l

⇒ W' = - 0.15 x 0.3 x 9.8 x 0.86 x 10

⇒ W' = - 3.79 J

⇒ Net work done = W + W' = -3.79 - 3.79 = - 7.58 J

(iv) Which of the above forces (except applied force) is/are conservative forces?

Gravitational force

Note:

• Conservative force depends only upon the initial and the final position of the object it doesn't depend on the path covered by the object.
• The work done in an enclosed path is zero.