7) An experimental rocket plane lands and skids on a dry lake bed. If it's traveling at 76.0 m/s when it touches down, how far does it slide before coming to rest? Assume the coefficient of kinetic friction between the skids and the lake bed is 0.590.
8) Find the initial acceleration of a rocket if the astronauts onboard experience six times their normal weight during an initial vertical ascent. (In this exercise, the scale force is replaced by the normal force.)
m/s2
For question 7:
To find the distance the rocket plane slides before coming to rest, we can use the equation of motion for an object experiencing friction:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the plane comes to rest)
u = initial velocity (76.0 m/s)
a = acceleration (opposite in direction to the motion, due to friction)
s = distance
Since we are given the initial velocity, final velocity, and the coefficient of kinetic friction, we can rearrange the equation to solve for the distance:
s = (v^2 - u^2) / (2a)
The only unknown in this equation is 'a.' We can find 'a' using the equation for friction:
frictional force = coefficient of friction × normal force
The normal force is the force exerted by the ground on the rocket plane, which is equal in magnitude but opposite in direction to the weight of the rocket plane (mg).
The frictional force opposing the motion is given by:
frictional force = coefficient of friction × normal force = coefficient of friction × mg
Since the rocket plane is on the ground and not accelerating vertically, the normal force is equal to the weight of the plane. Therefore:
frictional force = coefficient of friction × weight of the plane = coefficient of friction × mg
The weight of the plane is given by mg where m is the mass of the rocket plane and g is the acceleration due to gravity.
Now we can find the acceleration 'a' using Newton's second law:
frictional force = mass × acceleration
So,
coefficient of friction × mg = m × a
Simplifying and canceling 'm' on both sides of the equation, we get:
a = coefficient of friction × g
Now we have found the value of 'a' and we can substitute it into the equation for distance:
s = (v^2 - u^2) / (2 × a)
Substituting the given values:
v = 0 m/s (final velocity)
u = 76.0 m/s (initial velocity)
a = coefficient of friction × g = 0.590 × 9.8 m/s^2 (acceleration due to gravity)
Solving for 's' will give us the distance the rocket plane slides before coming to rest.
For question 8:
To find the initial acceleration of the rocket, we can use the concept of apparent weight. The apparent weight experienced by the astronauts is the normal force acting on them, which is equal to their mass multiplied by their acceleration.
Given that the astronauts experience six times their normal weight during the initial vertical ascent, the apparent weight (normal force) is 6 times their actual weight.
Therefore, we can write:
apparent weight = normal force = 6 × actual weight
Using Newton's second law (F = m × a), we know that force is equal to mass multiplied by acceleration. In this case, the force is the apparent weight experienced by the astronauts, and the mass is the mass of the astronauts.
Therefore,
apparent weight = normal force = mass × acceleration
Substituting the given value of the apparent weight (6 × actual weight) for the apparent weight in the equation, we get:
6 × actual weight = mass × acceleration
Since the actual weight is given by mass × g (where g is the acceleration due to gravity), we can write:
6 × (mass × g) = mass × acceleration
Cancelling out 'mass' on both sides of the equation, we get:
6 × g = acceleration
So, the initial acceleration of the rocket is 6 times the acceleration due to gravity.