Question

0•2 kg mass of body suspended by 1 meter length of string if we displaced the body by using force at which the string make 30. Degree angle with horigontal then find the work done against with gravity

Answer 1

Answer on Question #45842-Physics-Mechanics-Kinematics-Dynamics

$m = 0.2\,kg$ mass of body suspended by $l = 1$ meter length of string if we displaced the body by using force at which the string make $\theta = 30$ Degree angle with horizontal then find the work done against with gravity.

Solution

The work done against with gravity is the difference between final and initial gravitational potential energy:

$$W = U_f - U_i.$$

Let initial gravitational potential energy be

$$U_i = -mgl,$$

where $g = 10\frac{m}{s^2}$ is the acceleration due to the gravity.

Then final gravitational potential energy is

$$U_f = -mgl \sin \theta.$$

The work done against with gravity is

$$W = U_f - U_i = -mgl \sin \theta - (-mgl) = mgl(1 - \sin \theta) = 0.2 \cdot 10 \cdot 1(1 - \sin 30{}^\circ) = 1J.$$

Answer: 1 J.

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