Question

Dwight uses a spring (k = 70 N/m) on a horizontal surface to make a launcher for model cars. The spring is attached to a holder that has a mass of 0.3 kg. If he compresses the spring and releases it, the launcher will move back and forth in periodic motion.
a. If he compresses the spring by 0.4 m and releases it, what is the acceleration of the launcher as the spring reaches its natural length?

b. What is the maximum kinetic energy fo the launcher?

c. What is the maximum speed of the launcher?

Answer 1

Part A)

As we know that spring force is given by

F = kx

here x = stretch in the spring from natural length

So here  when spring reaches to its natural length

Force due to spring = 0

so acceleration = 0

Part b)

When spring is compressed from its natural length it will have elastic potential energy in it

so it is given by

$U = \frac{1}{2}kx^2$

now we know that there is no friction in it so maximum kinetic energy of the launcher must be equal to the elastic potential energy of the spring

$KE = \frac{1}{2}kx^2$

here we have

k = 70 N/m

x = 0.4 m

$KE = \frac{1}{2}(70)(0.4)^2$

$KE = 5.6 J$

Part c)

Now to find the speed we know that

$KE = \frac{1}{2} mv^2$

$5.6 = \frac{1}{2}0.3v^2$

$v = 6.11 m/s$

so its speed is 6.11 m/s

Answer 2

a) In the very beginning its acceleration is quite big, but when the spring is stretched to the normal length, its acceleration is 0.

b) Speed will be

KE=1/2 mv^2=1/2 kx^2=5.6 J

c) $v=\sqrt{\frac{2KE}{m} }=6.11$ m/s