Question

A force $F=10\sqrt{2}N$ acts an angle of ${45}^o$ above the horizontal on a $2kg$ block placed on a rough horizontal surface. The coefficient of friction between the block and surface is $0.2$ Find the work done by the force $F$ on the block in $5s$ initially the block is at rest. [Take $g=10/s^2$]

• A. $250J$
• B. $2500J$
• C. $500J$
• D. $50J$

Answer 1

The correct option is D $50J$

$Fx=10N\left(F\cos {45}^o\right)$

$Fy=10N\left(F\sin {45}^o\right)$

$f_{max}=\mu N$

$=\mu (Mg-Fy)$

$=0.2(20-10)$

$=2N$

So, $a=Fx-f/m=10-2/2=4m/{\sec }^2$

By wrong emergency theorem :-

$F.x+fx=△xE$

$(F.x)+(-2\times 5)=\frac{1}{2}mV+2-\frac{1}{2}mV{i}^2$

$(F.x)-10=\frac{1}{2}m(V{i}^6+2ax-V{i}^2)$

$F.x-10=1(2\times \frac{8}{2}\times 5)$

$F.x=500$

Option $\left(d\right)$