A body of mass m dropped from a height h reac | vedclass

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Question

(A) According to the Work-Energy Theorem,the total work done on the body is equal to the change in its kinetic energy.$W_{total} = \Delta K$$W_{gravit

Vedclass · Accepted Answer

$-0.68 mgh$

Vedclass · Answer

(A) According to the Work-Energy Theorem,the total work done on the body is equal to the change in its kinetic energy.$W_{total} = \Delta K$$W_{gravity} + W_{air\ friction} = K_f - K_i$Given that the body is dropped,initial velocity $u = 0$,so $K_i = 0$.The work done by gravity is $W_{gravity} = mgh$.The final kinetic energy is $K_f = \frac{1}{2} m v^2 = \frac{1}{2} m (0.8 \sqrt{gh})^2$.$K_f = \frac{1}{2} m (0.64 gh) = 0.32 mgh$.Substituting these values into the theorem:$mgh + W_{air\ friction} = 0.32 mgh - 0$$W_{air\ friction} = 0.32 mgh - mgh = -0.68 mgh$.Thus,the work done by air friction is $-0.68 mgh$.

$A$ body of mass $m$ dropped from a height $h$ reaches the ground with a speed of $0.8 \sqrt{gh}$. The value of work done by the air friction is $.....$

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