A ball with a mass of #144 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #96 (kg)/s^2# and was compressed by #4/3 m# when the ball was released. How high will the ball go?

Answer 1

The height is #=60.5m#

The spring constant is #k=96kgs^-2#

The compression is #x=4/3m#

The potential energy is

#PE=1/2*96*(4/3)^2=85.33J#

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=0.144kg#

#PE_(ball)=85.33=0.144*9.8*h#

#h=85.33*1/(0.144*9.8)#

#=60.5m#

The height is #=60.5m#

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the maximum height the ball will reach, we can use the conservation of mechanical energy principle. The potential energy stored in the spring when compressed is equal to the kinetic energy of the ball at its maximum height.

The potential energy stored in the spring is given by the formula (PE = \frac{1}{2} kx^2), where (k) is the spring constant and (x) is the compression of the spring.

The kinetic energy of the ball at its maximum height is given by the formula (KE = \frac{1}{2} mv^2), where (m) is the mass of the ball and (v) is the velocity of the ball at its maximum height (which is 0 since the ball momentarily stops at its maximum height).

Equating the potential energy to the kinetic energy, we have:

(\frac{1}{2} kx^2 = \frac{1}{2} mv^2)

Solving for (x), we get:

(x = \sqrt{\frac{mv^2}{k}})

Given that (m = 144 , \text{g} = 0.144 , \text{kg}), (k = 96 , \text{(kg)/s}^2), and (x = \frac{4}{3} , \text{m}), we can solve for (v).

After finding (v), we can use the kinematic equation to find the maximum height:

(v^2 = u^2 - 2gh)

Where (u) is the initial velocity (0 in this case), (g) is the acceleration due to gravity (9.8 m/s(^2)), and (h) is the maximum height.

Solving for (h), we find:

(h = \frac{v^2}{2g})

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7