P= mv
p=(.5)(20)
p=10 kg/ms
Answer:
19.9 N/m
Explanation:
From the question,
Applying Hook's law
F = Ke.................. Equation 1
Where F = Force on the spring, k = spring constant, e = extension
But the force on the spring is the weight of the mass
Therefore,
mg = ke.................. Equation 2
Where m = mass. g = acceleration due to gravity
make e the subject of the equation
e = mg/e................ Equation 3
Given: m = 455 g = 0.455 kg, e = 22.4 cm = 0.224 m,
Constant: g = 9.8 m/s²
Substitute these values into equation 3
e = (0.455×9.8)/0.224
e = 19.9 N/m
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
