Atruck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. how strong must the rope be? μk = 0.373

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration

at: truck acceleration)

 equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 , t = final time =5 s

We replaced the known information in the equation(1):

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

mc = Car mass = 2230kg

g = Gravity acceleration=9.8

Normal force calculation:

Newton's first law

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N  Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

Calculation of the tension force in the rope (T):

Newton's Second law

T=10084,21 N

The rope must have a force of 10084,21 N

volts times amps 6 x 0.5 3w

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.