Question

A block of mass 2 kg is suspended by a weightless string of length 1m. A horizontal force is applied to displace the block slowly until the string makes an angle of 30 degree with the initial vertical direction. What is the work done by the applied force?​

Answer 1

Answer:

Work done = force * displacement.

Work will be MgL(1 - cos30)

= 20( 1 - √3/2)

W = 20(1 - √3/2)

Option A)

Answer 2

The work done is $20(1-\dfrac{\sqrt{3}}{2})\ J$

(A) is correct option.

Explanation:

Given that,

Mass of block = 2 kg

Length of string = 1 m

Angle = 30°

We need to calculate the force

Work done by tension + Work done by force (applied) + Work done by gravitational force = change in kinetic energy

Here, work done by tension is zero.

Put the value into the formula

$0+F\times AB-mg\times AC=0$

$F=mg\times(\dfrac{AC}{AB})$....(I)

According to figure,

$AB=l\sin30$

$AB=\dfrac{l}{2}$

$AC=OC-OA$

$AC=l-l\cos30$

$AC=l-\dfrac{l\sqrt{3}}{2}$

Put the value of AB and AC in equation (I)

$F=mg\times(\dfrac{l-\dfrac{l\sqrt{3}}{2}}{\dfrac{l}{2}})$

$F=2\times10\times(1-\dfrac{\sqrt{3}}{2})\times 2$

$F=40(1-\dfrac{\sqrt{3}}{2})\ N$

We need to calculate the work done

Using formula of work done

$W= F\cdot d$

$W=F\times AB$

$W=40(1-\dfrac{\sqrt{3}}{2})\times\dfrac{1}{2}$

$W=20(1-\dfrac{\sqrt{3}}{2})\ J$

Hence, The work done is $20(1-\dfrac{\sqrt{3}}{2})\ J$

Learn more :

Topic : Simple harmonic motion

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