Answer
$v_2=5.94m/s$, $N=8.529N$
Work Step by Step
The required speed and the normal reaction can be determined as follows:
According to the equation of conservation of energy
$\frac{1}{2}mv_1^+\frac{1}{2}kx_1^2+mgh_1=\frac{1}{2}mv_2^2+\frac{1}{2kx_2^2}+mgh_2$
We plug in the known values to obtain:
$0+(1500)(0.1)^2-2(0.3)(9.81)(1.5)=0.3v_2^2+0-2(0.3)(9.81)(1.5cos60)$
This simplifies to:
$v_2=5.94m/s$
We know that
$\Sigma F_n=m\frac{mv_2^2}{\rho}$
$\implies N-Wcos60=m\frac{v_2^2}{\rho}$
$\implies N=m(gcos60+\frac{v_2^2}{\rho})$
We plug in the known values to obtain:
$N=0.3(9.81cos60+\frac{(5.94)^2}{1.5})$
This simplifies to:
$N=8.529N$
