Question

If force F is acting always tangential to the path, and the equation of the path AB is $x^2=8y$ and the path BC is straight line, which is tangent on curve AB at point B. If coefficient of friction $\mu$ between block and path ABC is 0.5, the work done by force F to move block of mass 2 $kg$ from A to C very slowly is $76aJ$. Then the value of $a$ is . [Take g = 10 $m/s^2$]. Write upto two digits after the decimal point.

Answer 1

Slope of line BC is slope of curve AB at B.

$Slope=\frac{dy}{dx}=\frac{2x}{8}=\frac{2\times 4}{8}=1\Rightarrow \theta ={45}^{\circ }$
As we have slope of BC and BC length we can calculate lengths of sides of the triangle formed.
If mass m is taken from A to C slowly work done by friction will always be equal to the $W_f=-\mu mgx$
Now,
$W_{net}=\Delta KE=0$
$W_{net}=W_F-\Delta PE+W_f$
$W_{net}=W_F-mgh-\mu mgx$
$W_{net}=W_F-mg(10+2)-\mu mg(10+4)$
$\Rightarrow W_F-\left(2\right)\left(10\right)\left(12\right)-\left(0.5\right)\left(2\right)\left(10\right)\left(14\right)=0$
$\Rightarrow W_F-240-140=0$
$\Rightarrow W_F=380J=76\times 5J\Rightarrow a=5.$