A ball with a mass of 350 g350g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 16 (kg)/s^216kgs2 and was compressed by 7/4 m74m when the ball was released. How high will the ball go?

3 Answers
Apr 11, 2018

The height reached by the ball is =7.14m=7.14m

Explanation:

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The spring constant is k=16kgs^-2k=16kgs2

The compression of the spring is x=7/4mx=74m

The potential energy stored in the spring is

PE=1/2kx^2=1/2*16*(7/4)^2=24.5JPE=12kx2=1216(74)2=24.5J

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^2KEball=12mu2

Let the height of the ball be =h =h

The acceleration due to gravity is g=9.8ms^-2g=9.8ms2

Then ,

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

Mass of the ball is m=0.350kgm=0.350kg

PE_(ball)=24.5=0.350*9.8*hPEball=24.5=0.3509.8h

h=24.5*1/(0.350*9.8)h=24.510.3509.8

=7.14m=7.14m

The height reached by the ball is =7.14m=7.14m

Apr 11, 2018

First we have to find the force of the spring.

Explanation:

Using the spring constant formula this can be found

F = -kxF=kx

F = 16 xx 7/4F=16×74

F = 28NF=28N

Then the acceleration is:

a = F/ma=Fm

a = 28/0.35a=280.35

a = 80a=80 ms^-2ms2

To find the velocity at which the ball leaves the spring the following formula can be used:

v^2 = u^2 + 2axv2=u2+2ax

v^2 = 0 + 2 xx 80 xx 7/4v2=0+2×80×74

v^2 = 280v2=280

v = 16.73v=16.73 ms^-1ms1

Now this is a projectile motion question.

v^2 =u^2 + 2aHv2=u2+2aH

0 = 16.73^2 + 2 xx- 9.8 xx H0=16.732+2×9.8×H

H = 280/(2 xx 9.8)H=2802×9.8

H = 14.29H=14.29 mm

The ball travels 14.2914.29 mm high.

Apr 11, 2018

approx 7 m7m

Explanation:

I will assume that all of the potential energy in the spring will turn into kinetic energy and the only thing opposing the motion of the ball is the acceleration of free fall acting in the opposite direction of it's motion.

Now, the potential energy in the spring is given by:

E_p=0.5kDeltax^2
k=spring constant
x=compression of the spring

E_p=0.5 times (16) times (7/4)^2=24.5 J

E_p=E_k

E_k=0.5 times m times v^2

24.5=0.5 times 0.35 times v^2

v=11.71 ms^-1 (the initial velocity of the ball)

We know that our final velocity will be 0 as this is the point where the ball will start falling down again due to the acceleration of free fall acting in the opposite direction.

Using the kinematic equation:
v=u+at
v=final velocity (0)
u=inital velocity
a=acceleration (-9.81, as gravity is opposing or direction of motion)
t=time

0=11.71-9.81t
t=1.193 s
So it takes 1.193 seconds for the ball to reach its maximum height.

Now using another kinematic equation to find the distance travelled:

s=((v+u)t)/(2)
s=distance travlled
v=final velocity (0)
u=inital velocity- 11.71
t=time

s=(1.193 times 11.71)/2
s=6.98 m approx 7m