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Two stacked boxes are sliding along an ice rink with an initial velocity vo to the right. You are trying to stop the boxes so you exert a force F'y1 on the top box as indicated in the figure. Y1 1 The magnitude of the force that you exert is variable, and follows the equation: Fy1 = Foe bt with Fo and b constants, and t the time since you began exerting the force. Because the boxes are sliding on ice, the friction between box and the ground is negligible. There is friction between the two boxes, and they are observed to accelerate together as you exert the force to slow them down. Note: This problem will be much easier if you solve everything symbolically before you substitute numerical values for the parameters. Use the following values for the parameters: mị = 6.7 kg mą = 2.5 kg = 23.4° Fo = 100 N b = 0.16 s-1 ti = 2.3 s vo = 9.1 m/s

Two stacked boxes are sliding along an ice rink with an initial velocity vo to the right. You are trying to stop the boxes so you exert a force F'y1 on the top box as indicated in the figure. Y1 1 The magnitude of the force that you exert is variable, and follows the equation: Fy1 = Foe bt with Fo and b constants, and t the time since you began exerting the force. Because the boxes are sliding on ice, the friction between box and the ground is negligible. There is friction between the two boxes, and they are observed to accelerate together as you exert the force to slow them down. Note: This problem will be much easier if you solve everything symbolically before you substitute numerical values for the parameters. Use the following values for the parameters: mị = 6.7 kg mą = 2.5 kg = 23.4° Fo = 100 N b = 0.16 s-1 ti = 2.3 s vo = 9.1 m/s

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Two stacked boxes are sliding along an ice rink with an initial velocity v→0 to the right. You are trying to stop the boxes so you exert a force F→Y1 on the top box as indicated in the figure.

The magnitude of the force that you exert is variable, and follows the equation: FY1=F0e^(−bt) with F0 and b constants, and t the time since you began exerting the force.

Because the boxes are sliding on ice, the friction between box 2 and the ground is negligible. There is friction between the two boxes, and they are observed to accelerate together as you exert the force to slow them down.

This problem will be much easier if you solve everything symbolically before you substitute numerical values for the parameters.

Use the following values for the parameters:

m1=6.7 kg
m2=2.5 kg
ϕ=23.4∘
F0=100 N
b=0.16 s−1
t1=2.3 s

v0=9.1 m/s

I calculated acceleration to be a=2.99 m/s^2. I am trying to find the speed of the boxes at time 1 (t1=2.3 s), the friction force of box one on box two, and the stopping time. 

To calculate speed I think I need to be taking an integral of the accleration. to find the friction force I think I need to use Newtons second law, and I cant figure out what I need to do for finding the stopping time. Can you please help me through this?

Two stacked boxes are sliding along an ice rink with an initial velocity vo to the right. You are trying to stop the boxes so you exert a force Fy1 on the top box as indicated in the figure. F Y1 1 2 The magnitude of the force that you exert is variable, and follows the equation: Fy1 = F e bt with Fo and b constants, and t the time since you began exerting the force. Because the boxes are sliding on ice, the friction between box 2 and the ground is negligible. There is friction between the two boxes, and they are observed to accelerate together as you exert the force to slow them down. Note: This problem will be much easier if you solve everything symbolically before you substitute numerical values for the parameters. Use the following values for the parameters: m1 = 6.7 kg m2 = 2.5 kg $ = 23.4° Fo = 100 N b = 0.16 s1 ti = 2.3 s vo = 9.1 m/s
Transcribed Image Text:Two stacked boxes are sliding along an ice rink with an initial velocity vo to the right. You are trying to stop the boxes so you exert a force Fy1 on the top box as indicated in the figure. F Y1 1 2 The magnitude of the force that you exert is variable, and follows the equation: Fy1 = F e bt with Fo and b constants, and t the time since you began exerting the force. Because the boxes are sliding on ice, the friction between box 2 and the ground is negligible. There is friction between the two boxes, and they are observed to accelerate together as you exert the force to slow them down. Note: This problem will be much easier if you solve everything symbolically before you substitute numerical values for the parameters. Use the following values for the parameters: m1 = 6.7 kg m2 = 2.5 kg $ = 23.4° Fo = 100 N b = 0.16 s1 ti = 2.3 s vo = 9.1 m/s
a(tı) 2.99 m/s^2
Transcribed Image Text:a(tı) 2.99 m/s^2
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