Question

A 0.300-kg cart connected to a light spring for which the force constant is 18.0 N/m oscillates on a frictionless, horizontal air track. Use an energy approach to respond to the questions below. (A) Calculate the maximum speed of the cart if the amplitude of the motion is 6.00 cm. (B) What is the velocity of the cart when the position is 3.00 cm?

Answer 1

Let's ,the spring constant is K =18.0 N/m

and M=0.300 kg

and A=6.0 cm =0.06 m

we know the formula of

ME =PE + KE.............................(1)

where ME= mechanical energy

KE= kinetic energy

PE= potential energy

  $\frac{1}{2}K.A=\frac{1}{2} MV^{2}+\frac{1}{2}KX^{2}....................(2)$

a) for V_{max ,}  X=0  

then the equation is =

  $\frac{1}{2}K.A=\frac{1}{2} MV_{max}^{2}$

  

  

КА 2 таr м

  $V_{max}^{2}=\frac{18*0.06}{0.300}$

  the maximum speed is 1.897 m/s

b) the velocity when x=3 cm=0.03 m

$V(x)=V_{max}\sqrt{1 - \frac{x^{2}}{A^{2}}}$

  $V(x=0.03)=1.897\sqrt{1 - \frac{(0.03)^{2}}{(0.06)^{2}}}$

$V(x=0.03)=1.897\sqrt{0.75}$

the velocity is 1.64 m/s when the cart position in 3 cm