A).
The tension in the rope when box is at rest is [tex]\fbox{\begin\\0\text{ N}\end{minispace}}[/tex].
Further Explanation:
The box, is tied to a horizontal rope, is placed on the surface of frictionless ice. The box is at rest and therefore, the velocity and acceleration of the box are zero.
Given:
The mass of the box, placed on the surface of frictionless ice, is [tex]50\text{ kg}[/tex].
The velocity of the box is [tex]0\text{ m/s}[/tex].
Concept:
The net force on the box, tied to the rope and placed on the surface of the frictionless ice, is equal to the product of mass of the box and the acceleration of the box.
The net force on the box is:
[tex]F=ma[/tex] ...... (1)
Here, [tex]m[/tex] is the mass of the box, [tex]a[/tex] is the acceleration of the box and [tex]F[/tex] is the net force on the box.
The box is placed on the surface of ice which is frictionless. Therefore, the net force on the box is equal to the tension produced in the rope.
The tension produced in the rope:
[tex]\fbox{\begin\\T=ma\end{minispace}}[/tex] ...... (2)
Here, [tex]T[/tex] is the tension produced in the rope.
Calculation:
Substitute the values in the equation (2).
[tex]\begin{aligned}\\T&=50\text{kg}\times0\\&=0\text{ N}\end{aligned}[/tex]
Thus, the tension in the rope when box is at rest is [tex]\fbox{\begin\\0\text{ N}\end{minispace}}[/tex].
B).
The tension in the rope when box is moving at a steady speed of [tex]5\text{ m/s}[/tex] is [tex]\fbox{\begin\\0\text{ N}\end{minispace}}[/tex].
Further Explanation:
The box is moving at the steady speed of [tex]5\text{ m/s}[/tex]. The acceleration of the box is zero because the rate of change of velocity with respect to time is zero.
Given:
The velocity of the box is [tex]5\text{ m/s}[/tex].
Calculation:
Substitute the values in the equation (2).
[tex]\begin{aligned}\\T&=50\text{kg}\times0\\&=0\text{ N}\end{aligned}[/tex]
Thus, the tension in the rope when box is moving at a steady speed of [tex]5\text{ m/s}[/tex] is [tex]\fbox{\begin\\0\text{ N}\end{minispace}}[/tex].
C).
The tension produced in the rope when the box is moving with velocity [tex]5\text{ m/s}[/tex] and it has acceleration [tex]5\text{ m}/\text{s}^2[/tex] is [tex]\fbox{\begin\\250\text{ N}\end{minispace}}[/tex].
Given:
The acceleration of the box is [tex]5\text{ m}/\text{s}^2[/tex].
The velocity of the box is [tex]5\text{ m/s}[/tex].
Calculation:
Substitute the values in the equation (2).
[tex]\begin{aligned}T&=50\text{ kg}\times5\text{ m}/\text{s}^2\\&=250\text{ N}\end{aligned}[/tex]
Thus, the tension produced in the rope when the box is moving with velocity [tex]5\text{ m/s}[/tex] and it has acceleration [tex]5\text{ m}/\text{s}^2[/tex] is [tex]\fbox{\begin\\250\text{ N}\end{minispace}}[/tex].
Learn more:
1. Conservation of energy brainly.com/question/3943029
2. The motion of a body under friction brainly.com/question/4033012
3. A ball falling under the acceleration due to gravity brainly.com/question/10934170
Answer Details:
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords:
Acceleration, frictionless surface, steady speed, tension in the rope, tension in the string, net force, ice-box system, 250N, 250 newton, 250 Newton, 0 N, 0 newton, 0 Newton, rope mass system, string mass system.
