A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl
1 answer:
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Answer:
1.0752 kgm/s
Explanation:
Considering when the drop was dropped from rest from a height,
mass of the ball, m = 0.120 kg
height, h = - 1.25 m
the initial velocity, u = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
Substituting the values,
V = ± 4.95 m/s
V = - 4.95 m/s
Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s
Considering when the ball rebounds from the floor,
assume the mass of the ball still remain, m = 0.120 kg
height, h = 0.820 m
the final velocity, v = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
Substituting the values,
U = ± 4.01 m/s
U = + 4.01 m/s
Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s
From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.
⇒ Impulse = Change in momentum
To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,
∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )
F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )
F.t = 0.120 kg × 8.96 m/s
Impulse = 1.0752 kgm/s
The impulse given to the ball by the floor is 1.0752 kgm/s
Answer:
a)
b)
Explanation:
In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).
After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:
We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:
When solving for the normal force we get:
we know that
W=mg
and
so after substituting we get that
N=F sin θ +mg
We also know that the kinetic friction is defined to be:
so we can find the kinetic friction by substituting for N, so we get:
Now we can find the sum of forces in x:
so after analyzing the diagram we can build our sum of forces to be:
we know that:
so we can substitute the equations we already have in the sum of forces on x so we get:
so now we can solve for the force, we start by distributing so we get:
we add to both sides so we get:
Nos we factor F so we get:
and now we divide both sides of the equation into so we get:
which is our answer to part a.
Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:
and
F cos θ = f
when substituting one into the other we get:
which can be solved for the static friction coefficient so we get:
which is the answer to part b.
The gravitational force between two objects is given by:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the first object has a mass of
, while the second "object" is the Earth, with mass 
. The distance of the object from the Earth's center is 
; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the first object has a mass of