A ball with a mass of #120 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #15 (kg)/s^2# and was compressed by #3/5 m# when the ball was released. How high will the ball go?

2 Answers
Jul 10, 2018

Answer:

The height reached by the ball is #=2.30m#

Explanation:

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The spring constant is #k=15kgs^-2#

The compression of the spring is #x=3/5m#

The potential energy stored in the spring is

#PE=1/2kx^2=1/2*15*(3/5)^2=2.7J#

This potential energy will be converted to kinetic energy when the spring is released and to potential energy of the ball

#KE_(ball)=1/2m u^2#

Let the height of the ball be #=h #

The acceleration due to gravity is #g=9.8ms^-2#

Then ,

The potential energy of the ball is #PE_(ball)=mgh#

Mass of the ball is #m=0.120kg#

#PE_(ball)=2.7=0.120*9.8*h#

#h=2.7/(0.120*9.8)#

#=2.30m#

The height reached by the ball is #=2.30m#

Answer:

#2.293\ m#

Explanation:

Assuming spring to be perfectly elastic, when the ball is released, the entire elastic energy of spring #1/2kx^2# is transferred to the ball in form of kinetic energy #1/2mv^2#
Where, #k# is spring constant,

#x# initial compression of spring

#m# is mass of ball

#u# is the velocity of ball when it is thrown vertical upward

hence, we have

#\text{K.E. of ball}=\text{elastic energy of spring}#

#1/2(0.120)u^2=1/2(15)(3/5)^2#

#u^2=135/3#

#u=\sqrt{135/3}#

Now, using third equation of motion:

#v^2=u^2+2as#

For the motion against gravity, we have #a=-g=-9.81\ \text{m/s}^2#, initial velocity #u=\sqrt{135/3}# & final velocity #v# at the maximum height #s=h# becomes zero i.e. #v=0#

hence, we have

#0^2=(\sqrt{135/3})-2(9.81)h#

#h=\frac{135}{3\cdot 19.62}#

#=2.293\ m#