Question

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

• A. 1947
• B. 1857
• C. 1792
• D. 1869

Answer 1

The correct option is C 1792

$X=2t^2+6t+4$
$V=4t+6$
at $t=0,V=6$
$t=4,V=$$4\times 4+6=22$
$W=\frac{1}{2}m(V_f^2-V_i^2)$
$=\frac{1}{2}.8({22}^2-6^2)=4(484-36)$
$=1792$