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Standard XII

Physics

K.E. in Relativistic Physics

A time varyin...

Question

A time varying net force acts on a $8.0 k g$ block. Particle's displacement is given by $X = A t^{2} + B t + C$ $(A = 2 m / s^{2}, B = 6 m / s, C = 4 m)$ . Find work done on the block in first $4 s$ of motion.

1947

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1857

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1792

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1869

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Solution

The correct option is C 1792
$X = 2 t^{2} + 6 t + 4$
$V = 4 t + 6$
at $t = 0, V = 6$
$t = 4, V =$ $4 \times 4 + 6 = 22$
$W = \frac{1}{2} m (V_{f}^{2} - V_{i}^{2})$
$= \frac{1}{2} .8 (22^{2} - 6^{2}) = 4 (484 - 36)$
$= 1792$

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A time varying net force acts on a 8.0 kg block. Particle's displacement is given by X=At2+Bt+C (A=2 m/s2,B=6 m/s,C=4 m). Find work done on the block in first 4 s of motion.

The correct option is C 1792X=2t2+6t+4 V=4t+6 at t=0,V=6 t=4,V=4×4+6=22 W=12m(V2f−V2i) =12.8(222−62)=4(484−36) =1792

A time varying net force acts on a $8.0 k g$ block. Particle's displacement is given by $X = A t^{2} + B t + C$ $(A = 2 m / s^{2}, B = 6 m / s, C = 4 m)$ . Find work done on the block in first $4 s$ of motion.

The correct option is C 1792
$X = 2 t^{2} + 6 t + 4$
$V = 4 t + 6$
at $t = 0, V = 6$
$t = 4, V =$ $4 \times 4 + 6 = 22$
$W = \frac{1}{2} m (V_{f}^{2} - V_{i}^{2})$
$= \frac{1}{2} .8 (22^{2} - 6^{2}) = 4 (484 - 36)$
$= 1792$