EXERCISE 4.9 (a) Suppose a hockey puck slides down a frictionless ramp with an acceleration of 5.00 m/s2. What angle
does the ramp make with respect to the horizontal? (b) If the ramp has a length of 6.00 m, how long does it take the
puck to reach the bottom? (c) Now suppose the mass of the puck is doubled. What’s the puck’s new acceleration down the
ramp?
(a) Well, the angle of the ramp must be really "acute" to have such a high acceleration! Badum tss! 🥁 In all seriousness though, we can use some trigonometry to find the angle. The acceleration, a, is equal to g * sin(theta), where g is the acceleration due to gravity. Plugging in the values, we get sin(theta) = a / g. Thus, the angle, theta, is equal to the inverse sine of (a / g).
(b) To find the time it takes for the puck to reach the bottom, we can use the kinematic equation: d = (1/2) * a * t^2, where d is the distance traveled. Plugging in the values, we have 6.00 m = (1/2) * 5.00 m/s^2 * t^2. Solving for t gives us t = sqrt(2*d/a).
(c) If the mass of the puck is doubled, we'll have to recalculate the acceleration. Remember that the acceleration down the ramp is determined by the component of gravity parallel to the ramp, which is given by g * sin(theta). Since the ramp angle doesn't change, the new acceleration will also be 5.00 m/s^2.
To find the angle of the ramp with respect to the horizontal, we can use the formula for the acceleration of an object sliding down an inclined plane:
a = g * sin(θ)
Where:
a = acceleration (given as 5.00 m/s^2)
g = acceleration due to gravity (approximately 9.8 m/s^2)
θ = angle of the ramp
Rearranging the formula, we get:
sin(θ) = a / g
Now we can solve for θ by taking the inverse sine (sin⁻¹) of both sides of the equation:
θ = sin⁻¹(a / g)
Plugging in the given values:
θ = sin⁻¹(5.00 / 9.8)
Using a calculator, we find that the angle of the ramp with respect to the horizontal is approximately 30.57 degrees.
To find the time it takes for the puck to reach the bottom of the ramp, we can use one of the kinematic equations:
s = ut + 0.5at^2
Where:
s = distance traveled (given as 6.00 m)
u = initial velocity (assumed to be 0 since the puck starts from rest)
a = acceleration (given as 5.00 m/s^2)
t = time
Rearranging the equation to solve for time, we get:
t = √(2s / a)
Plugging in the given values:
t = √(2 * 6.00 / 5.00)
Using a calculator, we find that it takes approximately 1.94 seconds for the puck to reach the bottom of the ramp.
To find the new acceleration of the puck when its mass is doubled, we can use Newton's second law of motion:
F = ma
Where:
F = force
m = mass
a = acceleration
In this case, we want to compare the acceleration of the puck before and after the mass is doubled. Since the force remains the same (no external forces are mentioned), we can write:
m1 * a1 = m2 * a2
Where:
m1 = initial mass (given)
a1 = initial acceleration (given as 5.00 m/s^2)
m2 = final mass (twice the initial mass)
a2 = final acceleration (unknown)
Simplifying the equation, we get:
a2 = (m1 * a1) / m2
Plugging in the given values:
a2 = (m1 * 5.00) / (2 * m1)
The mass cancels out, and we find that the new acceleration of the puck down the ramp is half of the initial acceleration, or 2.50 m/s^2.
adfdaf
(a) ma=mgsinα
sinα=a/g=5/9.8 =0.51
α=30.7°
(b) s=at²/2
t=sqrt(2s/a)=...
(c) the same acceleration