Answer:
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[tex]\frac{m}{s^{2} }[/tex]
at: truck acceleration[tex]\frac{m}{s^{2} }[/tex])
[tex]ac = at= \frac{vf-vi}{t-ti}[/tex] equation(1)
Known information:
vi = Initial speed = 0, ti = initial time = 0
vf = Final speed = 13 [tex]\frac{m}{s}[/tex], t = final time =5 s
We replaced the known information in the equation(1):
[tex]ac = at = \frac{13-0}{15-0} [/tex]
[tex]ac=ac=\frac{13}{15} \frac{m}{s}[/tex]
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:
[tex]Wc=mc*g[/tex]
mc = Car mass = 2230kg
g = Gravity acceleration=9.8 [tex]\frac{m}{s^{2} }[/tex]
[tex]Wc= 2230*9.8[/tex]
[tex]Wc=21854 N[/tex]
Normal force calculation:
Newton's first law
[tex]sum Fy= 0[/tex]
[tex]N-W=0[/tex]
[tex]N=W[/tex]
[tex]N=21854 N[/tex]
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:
[tex]Ff=0.373*21854[/tex]
[tex]Ff=8151.54 N[/tex]
Calculation of the tension force in the rope (T):
Newton's Second law
[tex] sum Fx= mc*ac[/tex]
[tex]T-Ff=mc*ac[/tex]
[tex]T=2230(\frac{13}{15}) + 8151.54[/tex]
T=10084,21 N
Answer: The rope must have a force of 10084,21 N
