Question

a person A of 50 kg rests on a swing of length 1 m makes an angle 37 with the vertical. another person B pushes him to swing on other side at 53° with vertical. the work done by B is......

Answer 1

hey mate....here is your answer.....

actually this answer was given by AdityaRocks1 ........he sent it to me......

Answer 2

Given:

Mass of person A, m1=50 kg

Length of swing, l=1 m

$\theta_1=37^{\circ}$

$\theta_2=53^{\circ}$

To find:

Work done by B

Solution:

$lcos\theta_1=1cos37^{\circ}=1\times \frac{4}{5}=\frac{4}{5}m$

$lcos\theta_2=1\times cos53^{\circ}=\frac{3}{5}m$

Now,

$h=lcos\theta_1-lcos\theta_2$

$h=\frac{4}{5}-\frac{3}{5}$

$h=\frac{4-3}{5}=\frac{1}{5}m$

Work done by B

$W=mgh$

Where $g=10 m/s^2$

$W=50\times 10\times \frac{1}{5}$

$W=100 J$

Hence, the work done by B=100 J