Question

Calculate work done to move a body of mass 10 kg along a smooth inclined plane (theta =30°) velocity through a distance of 10 m



Neet aspirant kya phir koi bhi answer de do yrr plzz .....​

Answer 1

We need to apply an external force, F=mgsinθ up the plane, while the body is in motion downwards.

⇒W=Fd=mgsinθd

Given d=10m,m=10Kg,θ=30∘,g=10m/s2,

W=10×10×sin30×10

=100×

1

2

×10

=500J........

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Answer 2

Answer:

$\huge\underline\pink{\tt Hey\:Mate}$

$\large\underline{\tt Ur\:answer\:is\:500J}$

Here the motion is not accelerated the resultant force parallel to the plan must be zero .

So,

$\large{\tt F - Mg sin30°=0}$

$\large{\tt F = Mg sin30° }$

$\large {\tt and\:d=10m}$

$\large{\tt W=Fdcos(theta)=(Mgsin30°)dcos0°}</p><p>$

$\large{\tt W = 10×10×\frac{1}{2}×10×1}$

$\huge\blue{\boxed{\tt W=500 J}}</p><p>$

Hope this will help uh!!

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