Answer
a) Speed $v_{2}=\sqrt{2gx+v_{1}^{2}-\frac{2fx}{m}-\frac{kx^{2}}{m}}=3.65$ $m/s$.
b) Acceleration $a=\frac{f-mg+kx}{m}=4.00$ $m/s^{2}$
Work Step by Step
a) It is known that the work done by all forced other than the gravitational force or elastic force equals the change in the total mechanical energy of the system. $K_{1} + U_{1} + W_{other} = K_{2} + U_{2}$.
So $\frac{mv_{1}^{2}}{2} + mgh_{1} - fx=\frac{mv_{2}^{2}}{2} + mgh_{2} + \frac{kx^{2}}{2}$. Where $f$ is a friction force, $x=h_{1}-h_{2}$ is a displacement from point 1 to point 2 and $k=1.06\times10^{4}$ N/m is a force constant.
Speed $v_{2}=\sqrt{2gx+v_{1}^{2}-\frac{2fx}{m}-\frac{kx^{2}}{m}}=3.65$ $m/s$.
b) It is known from the second Newton's law that $F_{total}=ma$. So $ma=f-w+F=f-mg+kx$, where $f$ is a friction force, $w$ is a weight (gravity force) and $F$ is a spring force.
Acceleration $a=\frac{f-mg+kx}{m}=4.00$ $m/s^{2}$
