Question

A 80 N crate Slides with the constant speed a distance of 5m downward along a rough slope that makes an angle of 30° with the horizontal The work done by the force of gravity is?
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Answer 1

Answer:

Let us first calculate the Force experienced by the crate while travelling down the rough slope.

If the angle made by the slope with the horizontal is θ, then the force experienced by the object while travelling down the slope is given as:

• mg sinθ

Now since the surface is rough, the crate will also experience frictional force which is μN. (where, N refers to the Normal which is nothing but mg.)

Hence the total net force by which the object comes down is:

⇒ mg sinθ - μN   ...(i)

(Since friction acts opposite to the force, we use negative sign.)

(I am assuming that the Force + Frictional Force = 80 N. If it's not the case, then you must use Eqn. (i) to get your answer.)

From the question we get that the Net Force after considering Friction and external factors is 80 N. Hence the component of Force at the angle 30° is:

⇒ Force = mg sinθ

⇒ Force = 80 × sin (30)

⇒ Force = 80 × 0.5 = 40 N.

Now we know that work done by an object is given as:

⇒ Work Done = Force × Displacement

Acc. to the question, Displacement = 5 m. Hence we get,

⇒ Work Done = ( 40 ) N × 5 m

⇒ Work Done = 200 J

Hence the work done by the force of gravity is 200 J.

Answer 2

Answer:

Given :-

• A 80 N crate slide with the constant speed a distance of 5 m downward along a rough slope that makes an angle of 30° with the horizontal.

To Find :-

• What is the work done by the force.

Formula Used :-

$\bigstar\: \: \boxed{\bold{\pink{Work\: Done =\: F\: sin\theta\: d}}}\: \: \bigstar$

where,

• F = Force
• $\sf \theta$ = Angle of the slope
• d = Distance

Solution :-

Given :

• Force of the slope = 80 N
• Downward distance along a rough slope = 5 m
• Angle of slope = 30°

According to the question by using the formula we get,

$\longrightarrow \sf Work\: Done =\: 80 \times sin\: 30^{\circ}\: \times 5$

As we know that :

$\leadsto \sf\bold{\purple{sin\: 30^{\circ} =\: \dfrac{1}{2}}}$

$\longrightarrow \sf Work\: Done =\: 80 \times \dfrac{1}{2} \times 5$

$\longrightarrow \sf Work\: Done =\: 80 \times \dfrac{1 \times 5}{2}$

$\longrightarrow \sf Work\: Done =\: {\cancel{80}} \times \dfrac{5}{\cancel{2}}$

$\longrightarrow \sf Work\: Done =\: 40 \times \dfrac{5}{1}$

$\longrightarrow \sf Work\: Done =\: \dfrac{40 \times 5}{1}$

$\longrightarrow \sf Work\: Done =\: \dfrac{200}{1}$

$\longrightarrow \sf\bold{\red{Work\: Done =\: 200\: J}}$

${\small{\bold{\underline{\therefore\: The\: work\: done\: by\: the\: force\: of\: gravity\: is\: \bold{\red{200\: J}}\: .}}}}$