A horizontal rope is tied to a 50 kg 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 5.0 m/s? c. The box has&nbsp;vx=5.0ms&nbsp;and&nbsp;ax=5.0ms2?

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A horizontal rope is tied to a 50 kg 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 5.0 m/s? c. The box has&amp;nbsp;vx=5.0ms&amp;nbsp;and&amp;nbsp;ax=5.0ms2?

Melinda Olson · Accepted Answer

Given values: m=50kg a) The box is at rest, which means that it does not move, the acceleration is zero if it is accelerated by zero, the the sum of all forces acting on the box is equal to zero, According to Newton&#039;s second law: F=ma F=m⋅(0ms2) F=0 N T1=0 N b) v=5ms The box mobes at a constant speed. The acceleration, in this case, is zero. F=ma F=m⋅(0ms2) F=0 N T2=0 N c) vx=5ms ax=5ms2 In this case the acceleration of the box is not zero. F=ma F=50kg⋅(5ms2) F=250 N T3=250 N a) T1=0 N b)T2=0 N c)T3=250 N

alenahelenash · Accepted Answer

Step 1:a. The box is at rest:In this case, since the box is at rest, there is no net force acting on it in the horizontal direction. Therefore, the tension in the rope is equal to zero (T=0).b. The box moves at a steady 5.0 m/s:When the box moves at a steady velocity, the net force acting on it must be zero (∑F=0). The only horizontal force acting on the box is the tension in the rope (T). According to Newton&#039;s second law, the net force is given by the product of mass and acceleration (∑F=m·a). Since the box is moving at a constant velocity, the acceleration is zero (a=0). Thus, we have:T−0=m·aT=m·aSubstituting the given values, we get:T=(50kg)·(0m/s2)T=0NTherefore, the tension in the rope when the box moves at a steady 5.0 m/s is zero (T=0).c. The box has vx=5.0m/s and ax=5.0m/s2:In this case, both the velocity and acceleration of the box are in the positive x-direction. The tension in the rope can be determined using Newton&#039;s second law as before:T−frictional&amp;nbsp;force=m·aHowever, in this scenario, we need to consider the presence of frictional force. Since the ice is frictionless, there is no frictional force acting on the box. Therefore, the equation becomes:T−0=m·aT=m·aStep 2:Substituting the given values, we have:T=(50kg)·(5.0m/s2)T=250NHence, the tension in the rope when the box has a velocity of 5.0m/s and an acceleration of 5.0m/s2 is 250N (T=250N).

A horizontal rope is tied to a 50 kg box on frictionless ice. What is the tensio

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