Answer:
a) To solve for the speed of the car, we can use the conservation of energy equation:
(1/2)mv^2 = mgh - (1/2)kx^2
where m is the mass of the car, v is the speed of the car, g is the acceleration due to gravity, h is the height of the car above the starting point, k is the spring constant, and x is the distance the spring is compressed.
Plugging in the given values, we get:
(1/2)(0.250 kg)v^2 = (0.250 kg)(9.81 m/s^2)(0.110 m) - (1/2)(450.0 N/m)(0.0600 m)^2
Solving for v, we get:
v = 1.23 m/s
Therefore, the speed of the car when it reaches 0.110 m above the starting point is 1.23 m/s.
b) To estimate the power expended by a child raising a 0.25-kg toy 0.75 m in 2.0 s, we can use the work-energy theorem:
W = ΔE = mgh
where W is the work done, m is the mass of the toy, g is the acceleration due to gravity, h is the height the toy is raised, and ΔE is the change in energy.
Plugging in the given values, we get:
W = (0.25 kg)(9.81 m/s^2)(0.75 m) = 1.84 J
Next, we can use the definition of power:
P = W/t
where P is power, W is work, and t is time.
Plugging in the given values, we get:
P = 1.84 J / 2.0 s = 0.92 W
Finally, we can convert watts to horsepower:
1 hp = 746 W
0.92 W = 0.92 / 746 hp = 0.0012 hp
Therefore, the power expended by a child raising a 0.25-kg toy 0.75 m in 2.0 s is approximately 0.0012 hp.