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### A body of mass 'm' dropped from a height 'h' ....

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:

(A) -0.68 mgh
(B) 0.64 mgh
(C) mgh
(D) 1.64 mgh

Solution

Short Method

Air friction force is resistive force whose work-done must be negative. The only option with negative value is (A).

Detailed Method

Work done by all forces $W_{all} = \Delta K$

$\therefore W_{mg} + W_{air friction} = \frac {1}{2} mv^2 - 0$

$\Rightarrow W_{fr} = \frac {1}{2} m \times (0.8 \sqrt {gh} )^2 - mgh$

$\Rightarrow W_{fr} = 0.32 mgh - mgh = -0.68 mgh$

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$