Question

A constant external force p = 160N is applied to a 20.0 kg box, which is on a rough horizontal surface. When the force pushes the box a distance of 8.00 m, the speed changes from 0.005 m/s. to 2.60 m/s. The work done by friction during this process is closest to:

• A. $+1212J$
• B. $+1110J$
• C. $+1170J$
• D. $+1040J$

Answer 1

The correct option is A $+1212J$

Acceleration of the block can be found using third equation of motion $v^2=u^2+2as$

$\Rightarrow {2.6}^2-{0.005}^2=2a\times 8$

$\Rightarrow a=\frac{{2.6}^2-{0.005}^2}{2\times 8}=$ $\frac{2.605\times 2.595}{16}m/s^2$

Also, $a=\frac{160-Forc{e}_{\left(Friction\right)}}{20}=$ $\frac{2.605\times 2.595}{16}m/s^2$

$\Rightarrow Friction=151.55N$

$\Rightarrow Work=F.S=1212J\left(approx\right)$