The range of force exerted at the end of the rope is 285.7 N to 1,000 N.
<h3>Net horizontal force of the cylinder</h3>
The net horizontal force of the cylinder when it is at equilibrium position is determined by applying Newton's second law of motion.
∑F = 0
F - μFn = 0
F - 0.2(5,000) = 0
F - 1,000 = 0
F = 1,000 N
The strength of the applied force increases as the number of turns of the rope increases.
minimum force = total force/number of turns of rope
minimum force = 1,000/3.5
minimum force = 285.7 N
Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.
Learn more about Newton's second law of motion here: brainly.com/question/3999427
Answer:
a) They are in the same point
b) t = 0 s, t = 2.27 s, t = 5.73 s
c) t = 1 s, t = 4.33 s
d) t = 2.67 s
Explanation:
Given equations are:
Constants are:
a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both 
and 
depend on t and don't have constant terms.
So both cars A and B are in the same point.
b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).
s, 
s
c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.
s, 
s
d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.
s
Given Information:
Mass = m = 500 kg
Acceleration = a = 10 cm/s²
Required Information:
Magnitude of rightward net force = F = ?
Answer:
Magnitude of rightward net force = 50 N
Explanation:
From the Newton's second law of motion
F = ma
Where m is the mass and a is the acceleration
To get force in Newtons first convert 10 cm/s² into m/s²
10/100 = 0.1 m/s²
F = 500*0.1
F = 50 N
Therefore, the magnitude of rightward net force acting on it is 50 Newtons.
Answer:
The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Explanation:
The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.
The angle between the 2 forces and displacement is ∅ = 15°.
First we have to calculate the work done by the individual force and then we can calculate the total work.
The work done on a particle by a constant force F during a straight line displacement s is given by following formula:
W = F*s
W = F*s*cos∅
With ∅ = the angles between F and s
The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N
The total work done can be calculated as followed:
Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2
Wtotal = 2Fs*cos∅
Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°
Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>
<u />
The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
