a 60kg rock in on a mountain in banff national park breaks free from a ledge and falls down 500 meters to the bottom of the moun
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The ball will decelerate as it moves upwards.
The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
The given parameters;
- initial velocity of the ball, u = 1.25 m/s
- time of motion of the ball, t = 4.22 s
As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.
Thus, the ball decelerate as it moves upwards.
The acceleration of the ball is calculate as;
<em>at the highest point on the incline plane, the final velocity </em><em> is zero</em>
Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
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(a) -2451 N
We can start by calculating the acceleration of the car. We have:
is the initial velocity
v = 0 is the final velocity of the car
d = 125 m is the stopping distance
So we can use the following equation
To find the acceleration of the car, a:
Now we can use Newton's second Law:
F = ma
where m = 1100 kg to find the force exerted on the car in order to stop it; we find:
and the negative sign means the force is in the opposite direction to the motion of the car.
(b)
We can use again the equation
To find the acceleration of the car. This time we have
is the initial velocity
v = 0 is the final velocity of the car
d = 2.0 m is the stopping distance
Substituting and solving for a,
So now we can find the force exerted on the car by using again Newton's second law:
As we can see, the force is much stronger than the force exerted in part a).
Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)