Question

the work done in moving a body of mass 5 kg from the bottom of a smooth incline plane to the top is 50j if the angle of inclination of the plane plane is 30° its length is

Answer 1

W=F×d=mgsin30 ×d =5×10×1/2×d =50;d=2m

Answer 2

Answer:

L = 2.03 m

Explanation:

Given,

• Mass of the body = m = 5 kg
• work done by the weight force  = w = 50 J
• Angel of inclination of the inclined plane = $\theta\ =\ 30^o$

Work done due to any force acting on the object is equal to the multiplication of the force acting on the object and the displacement covered by the object in the direction of the force applied.

Let L be the length of the inclined surface.

therefore the vertical component of the weight force = $w_y\ =\ mgsin30$

Work done due to the weight = W

$W\ =\ w_y\times L\\\Rightarrrow L\ =\ \dfrac{W}{mgsin30^o}\\\Rightarrow L\ =\ \dfrac{50}{5\times 9.81\times sin30^o}\\\Rightarorw L\ =\ 2.03\ m$

Hence the length of the inclined surface