Question 1

A spring-powered launcher has a spring with a constant of 50.0 N/cm. The launcher is prepared by compressing the spring by 5.00 cm. If the mass of the projectile is 10.0 grams, what is the speed of the projectile when it is launched?

Accepted Answer

The answer of A spring-powered launcher has a spring with...

Question 2

A 75 kg road bicycle racer rides down the highway from Pikes peak. During one stretch of the ride, he descends 5 meters in 12.5 seconds, while accelerating from 35 km/hr to 55 km/hr. Assuming rotational kinetic energy and dissipative forces to be negligible, what average power is supplied by the gravitational force during this time?

Accepted Answer

The average power supplied by the gravitational force can be calculated using the work-energy principle. The work done by gravity is equal to the change in kinetic energy plus the change in potential energy. First, convert speeds from km/hr to m/s: $35 \, \text{km/hr} = 9.72 \, \text{m/s}$ and $55 \, \text{km/hr} = 15.28 \, \text{m/s}$. The change in kinetic energy ($\Delta KE$) is $\frac{1}{2} m (v_f^2 - v_i^2)$, where $m = 75 \, \text{kg}$, $v_f = 15.28 \, \text{m/s}$, and $v_i = 9.72 \, \text{m/s}$. The change in potential energy ($\Delta PE$) is $mgh$, where $g = 9.8 \, \text{m/s}^2$ and $h = -5 \, \text{m}$ (negative because the racer is descending). The total work done by gravity is the sum of $\Delta KE$ and $\Delta PE$. Finally, power is work done over time, so divide the total work by the time interval (12.5 s). Calculating these values gives an average power of approximately 294 W.

Question 3

A spring-powered launcher has a spring with a constant of 150 N/cm. The launcher is prepared by compressing the spring by 5.00 cm. If the mass of the projectile is 5.00 grams, what is the speed of the projectile when it is launched?

Accepted Answer

The potential energy stored in the spring when it is compressed is given by:
$$U = \frac{1}{2}kx^2$$
where $k$ is the spring constant and $x$ is the compression distance. Substituting the given values:
$$U = \frac{1}{2}(150\,	ext{N/cm})(5.00\,	ext{cm})^2 = 1875\,	ext{J}$$
This potential energy is converted into kinetic energy of the projectile when the spring is released. The kinetic energy of an object with mass $m$ moving at speed $v$ is given by:
$$K = \frac{1}{2}mv^2$$
Since energy is conserved, we can equate the potential energy stored in the spring to the kinetic energy of the projectile:
$$K = U = 1875\,	ext{J}$$
Substituting the given mass of the projectile:
$$\frac{1}{2}(0.005\,	ext{kg})v^2 = 1875\,	ext{J}$$
Solving for $v$ gives:
$$v = \sqrt{\frac{2(1875\,	ext{J})}{0.005\,	ext{kg}}} \approx 86.6\,	ext{m/s}$$
Therefore, the correct answer is B.

Quiz 6: Conservation of Energy

A spring-powered launcher has a spring with a constant of 50.0 N/cm. The launcher is prepared by compressing the spring by 5.00 cm. If the mass of the projectile is 10.0 grams, what is the speed of the projectile when it is launched?

A spring-powered launcher has a spring with a constant of 150 N/cm. The launcher is prepared by compressing the spring by 5.00 cm. If the mass of the projectile is 5.00 grams, what is the speed of the projectile when it is launched?