Question

A box weighing 60N slides down from the top of an inclined plane making an angle of 300 with the horizontal if the body slides 4m along the plane, calculate the work done by the force of gravity.​

Answer 1

Given that: A box weighing 60 Newton's slides down from the top of an inclined plane making an angle of 300 with the horizontal, the body slides 4m along the plane.

To find: The work done by the force of gravity.

Solution: The work done by the force of gravity = 120√3 Joules

Using formula: Formula to find out the work done (according to the question) = Fs cosθ.

Where, F denotes force applied on any object, s denotes displacement and θ denotes the inclined plane making an angle.

Here, F is 60N, s is 4m and cosθ is √3/2° [∵ cos(30) = √3/2 (according to the trigonometry table)]

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \tt sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \tt cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \tt tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \tt \infty \\ \\ \tt cosec A & \tt \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \tt sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \tt \infty \\ \\ \tt cotA & \tt \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}$

Full Solution:

↪ Work = ️Fs cosθ

↪ Work = 60(4) √3/2

↪ Work = 240 √3/2

↪ Work = 120√3 Joules

↪ Hence, work done = 120√3 Joules.

Additional information:

♛ Work - If a body get displaced when a force acts on it, work it said to be done. Work is measured by the product of force and displacement of the body along the direction of force. If a body get displaced by S when a force F act's on it, then the work W = Fscosθ

[Where, ∅ denotes angle between force and displacement]

If both force and displacement are in the same direction, then W = FS.

Work is a scaler quantity and it's SI unit is joules.

♛ Energy = Capacity of doing work by a body is called it's energy. Energy is a scaler quantity & it's SI unit is joule Energy developed in a body due to work done on it is called mechanical energy. Mechanical energy is of two types –

• Potential energy
• Kinetic energy

Answer 2

Answer:

Work done is 120√3 J.

Step-by-step explanation:

Given :-

• Weight of the box = 60 N
• Angle made by the inclined plane = 30°
• Displacement made = 4 m

To find :-

Work done by the force of gravity.

Solution :-

➤ Work done = Fs cos θ

Where :

• F = Force
• s = Displacement
• θ = Angle made by the inclined plane

Substituting :

⇒ Work done = 60 × 4 × cos(30°)

⇒ Work done = 240 × cos(30°)

✮ cos(30°) = √3/2

⇒ Work done = 240 × √3/2

⇒ Work done = 120√3

∴ Work done is 120√3 J.