A horizontal rope is tied to a 45.0kg box on frictionless ice. What is the tension in the rope if:
A) The box is at rest?
B) The box moves at a steady vx = 4.40m/s ?
C) The box vx = 4.40m/s and ax = 4.60m/s2 ?
f=ma
a. zero, acceleration is zero
b.zero, acceleration iszero
c. tension=45*4.6
A) The tension in the rope when the box is at rest is 0. Don't worry, the box is not going anywhere. It's taking a well-deserved break on the frictionless ice.
B) When the box moves at a steady vx = 4.40 m/s, the tension in the rope is like the love handles of a chubby walrus—nonexistent! Again, no tension because the frictionless ice allows the box to slide smoothly.
C) Now things are getting interesting! When the box has a velocity vx = 4.40 m/s and acceleration ax = 4.60 m/s^2, the tension in the rope is like a magician's trick—it's going to pull a surprise on you. Calculate the net force on the box using F = ma, and you'll find that the tension in the rope will be equal to the mass of the box multiplied by its acceleration, which gives us (45.0 kg)(4.60 m/s^2) = 207 N. So, the tension in the rope is 207 Newtons, or the force it takes to bring the box to life and accelerate it forward.
To determine the tension in the rope, we need to consider the forces acting on the box. In this case, the only force acting on the box is the tension in the rope.
A) The box is at rest:
When the box is at rest, it means there is no acceleration, so the net force acting on the box is zero. Therefore, the tension in the rope must be equal to the weight of the box. The weight is given by the formula W = m * g, where m is the mass of the box and g is the acceleration due to gravity.
In this case, the weight of the box is W = 45.0 kg * 9.8 m/s^2 = 441 N. Therefore, the tension in the rope is also 441 N.
B) The box moves at a steady vx = 4.40 m/s:
When the box moves at a steady velocity, it means there is no acceleration. The tension in the rope must be equal to the frictional force acting on the box to keep it moving at a steady velocity. The frictional force is given by the formula F_friction = μ * N, where μ is the coefficient of friction and N is the normal force.
Since the box is on frictionless ice, there is no frictional force acting on it. Therefore, the tension in the rope is zero.
C) The box vx = 4.40 m/s and ax = 4.60 m/s^2:
When the box has a non-zero acceleration, we need to consider both the weight of the box and the net force acting on it. The net force is given by the formula F_net = m * a, where m is the mass of the box and a is the acceleration.
In this case, the net force is F_net = 45.0 kg * 4.60 m/s^2 = 207 N. Since there is no other force acting on the box, the tension in the rope must be equal to the net force. Therefore, the tension in the rope is 207 N.
24N
zero
0
2
5