A ball with a mass of #150 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #90 (kg)/s^2# and was compressed by #4/3 m# when the ball was released. How high will the ball go?
The height is
The spring constant is The compression is The potential energy in the spring is This potential energy will be converted to kinetic energy when the spring is released and to potential energy of the ball Let the height of the ball be Then , The potential energy of the ball is
By signing up, you agree to our Terms of Service and Privacy Policy
To find the maximum height reached by the ball, 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 when it is released.
The potential energy stored in the spring is given by (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 when it is released is given by (KE = \frac{1}{2} mv^2), where (m) is the mass of the ball and (v) is its velocity.
Since the spring is compressed by (4/3) m, we have (x = 4/3) m.
Using the given values:
(k = 90 , \text{(kg/s}^2))
(m = 0.150 , \text{kg})
We can find the velocity of the ball when it is released:
(PE = KE)
(\frac{1}{2} kx^2 = \frac{1}{2} mv^2)
(\frac{1}{2} \times 90 \times (4/3)^2 = \frac{1}{2} \times 0.150 \times v^2)
(v^2 = \frac{90 \times (4/3)^2}{0.150})
(v^2 = \frac{90 \times 16/9}{0.150})
(v^2 = \frac{90 \times 16}{0.150 \times 9})
(v^2 = \frac{1440}{1.35})
(v^2 = 1066.67)
(v = \sqrt{1066.67})
(v \approx 32.67 , \text{m/s})
Now that we have the velocity of the ball when it is released, we can use the kinematic equation to find the maximum height reached by the ball:
(v^2 = u^2 + 2as)
Where (v) is the final velocity (0 m/s at maximum height), (u) is the initial velocity (32.67 m/s), (a) is the acceleration due to gravity (-9.8 m/s(^2)), and (s) is the displacement (maximum height).
(0 = (32.67)^2 + 2 \times (-9.8) \times s)
(0 = 1067.91 - 19.6s)
(19.6s = 1067.91)
(s = \frac{1067.91}{19.6})
(s \approx 54.5 , \text{m})
Therefore, the ball will reach a maximum height of approximately 54.5 meters.
By signing up, you agree to our Terms of Service and Privacy Policy
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.
- An object with a mass of #3 kg# is hanging from an axle with a radius of #6 cm#. If the wheel attached to the axle has a radius of #2 cm#, how much force must be applied to the wheel to keep the object from falling?
- How much work does it take to push an object with a mass of #4 kg # up a #9 m # ramp, if the ramp has an incline of #(5pi)/12 # and a kinetic friction coefficient of # 8 #?
- If a spring has a constant of #7 (kg)/s^2#, how much work will it take to extend the spring by #38 cm #?
- A lifter takes 5 s to lift a 20-kg object by 10 m high. (1) What is the work done against gravity? (2) What is the power of the lifter?
- Why were simple machines invented?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7