Question

A $520$ gram block is sliding to the right on a surface that exerts a frictional force with coefficient of kinetic friction $\mu =0.400$. It collides with a horizontal spring with spring constant $k=18.0kg/s^2$ which compresses by $12.0cm$ as the block comes to rest. What was the initial speed of the block?

• A. $0.706m/s$
• B. $0.970m/s$
• C. $1.20m/s$
• D. $1.44m/s$
• E. Not enough information

Answer 1

Answer: (C)

$1.20m/s$

Given : $\mu =0.4$ $x=12$ cm $=0.12$ m $m=0.52$ kg $k=18.0$ $kg/s^2$

Let the initial speed of the block be $v$.

Final kinetic energy of the block is zero i.e $K.E_f=0$ J

Work done by spring force $W_{sp}=-\frac{1}{2}k{x}^2=-\frac{1}{2}\times 18.0\times (0.12{)}^2=-0.13$ J

Work done by friction force $W_f=-\mu \left(mg\right)x=-0.4\times (0.52\times 9.8)\times 0.12=-0.25$ J

Using work-energy theorem : $W_{sp}+W_f=K.E_f-K.E_i=0-\frac{1}{2}m{v}^2$

$\therefore$ $-0.13-0.25=-\frac{1}{2}\times \left(0.52\right)v^2$ $⟹v=\sqrt{1.46}=1.20$ $m/s$