A 57 kg boy and a 41 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleratio
1 answer:
Answer:
Acceleration of the boy a₁:
Explanation:
Conceptual analysis
We apply Newton's second law to the boy and the girl:
F = m*a (Formula 1)
F : Force in Newtons (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Nomenclature
m₁ : boy mass
m₂ : girl mass
a₁ : boy acceleration
a₂ : girl acceleration
F₁ : boy acceleration
F₂ : girl acceleration
Known data
m₁ = 57 kg
m₂ = 41 kg
a₂ = 2.6 m/s²
Problem development
We apply to Newton's third law of action and reaction, then:
F₁ = F₂ , We apply the formula (1):
m₁*a₁ = m₂*a₂
You might be interested in
1) The general equations of motion of the projectile on the x and y axis are:
where v0 is the initial velocity,
is the angle with respect to the ground, and 
is the gravitational acceleration. We can see that the motion of the projectile is an uniform motion on the x-axis and an uniformly accelerated motion on the y-axis.
First, we need to find what is the total horizontal displacement of the projectile when it is launched with an angle of
. To do that, we need to find first the time t at which the projectile lands to the ground, and we can find it by requiring y(t)=0:
that has two solutions: t=0 (beginning of the motion) and
and this is the time after which the projectile lands to the ground. If we substitute this value into the equation for x(t), we find the total horizontal displacement of the projectile:
with
.
If we call
the other angle at which the projectile reaches the same horizontal displacement, the total horizontal displacement in this case is
Since the horizontal displacement should be the same in the two cases, we can write x1=x2, which becomes:
Now let's remind that
so that we can rewrite the equation as
and using:
and we can see that there are two values of that satisfy the equation: 
and 
, which is the solution of our problem.
2) The vertical velocity of the ball at the very top of its trajectory is zero. In fact, the very top of the trajectory is the point where the ball starts to go down, so it means it is the moment when the the direction of the vertical velocity of the ball is changing from upward to downward, so it must be the moment when the vertical velocity is zero.
where v0 is the initial velocity,
First, we need to find what is the total horizontal displacement of the projectile when it is launched with an angle of
that has two solutions: t=0 (beginning of the motion) and
and this is the time after which the projectile lands to the ground. If we substitute this value into the equation for x(t), we find the total horizontal displacement of the projectile:
with
If we call
Since the horizontal displacement should be the same in the two cases, we can write x1=x2, which becomes:
Now let's remind that
and using
and we can see that there are two values of
2) The vertical velocity of the ball at the very top of its trajectory is zero. In fact, the very top of the trajectory is the point where the ball starts to go down, so it means it is the moment when the the direction of the vertical velocity of the ball is changing from upward to downward, so it must be the moment when the vertical velocity is zero.
Answer:
W = -1844.513 J
Explanation:
GIVEN DATA:
mass of spider man is m 74 kg
vertical displacement if spider is 11 m
final displacement = 11 cos 60.6 = - 6.753 m
change in displacement is = -6.753 - (-11) = 4.25 m
gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N
work done by gravity is
where 180 is the angle between spiderman weight and displacement
W = -1844.513 J