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下午9:08 9月24日周二 41%41 \% Back HW3-NL.pdf 5) A highway exit ramp is being designed...
Sep 25, 2024
Solution by Steps
step 1
First, we need to find the net force acting on the box. The box is pushed by two forces, F1 F_1 and F2 F_2 . We can resolve these forces into their x and y components. The x-component of F1 F_1 is given by F1x=F1cos(34) F_{1x} = F_1 \cos(34^\circ) and the y-component is F1y=F1sin(34) F_{1y} = F_1 \sin(34^\circ) . Similarly, for F2 F_2 , we have F2x=F2cos(20) F_{2x} = F_2 \cos(-20^\circ) and F2y=F2sin(20) F_{2y} = F_2 \sin(-20^\circ)
step 2
The net force in the x-direction can be expressed as: Fnet,x=F1x+F2x=120cos(34)+F2cos(20) F_{net,x} = F_{1x} + F_{2x} = 120 \cos(34^\circ) + F_2 \cos(-20^\circ) The net force in the y-direction is: Fnet,y=F1y+F2ymg=120sin(34)+F2sin(20)409.81 F_{net,y} = F_{1y} + F_{2y} - mg = 120 \sin(34^\circ) + F_2 \sin(-20^\circ) - 40 \cdot 9.81
step 3
Next, we calculate the acceleration of the box using the kinematic equation. The box moves 50.0 meters in 8.00 seconds, starting from rest. The acceleration a a can be calculated using: a=2dt2=250.0(8.00)2=100.064=1.5625m/s2 a = \frac{2d}{t^2} = \frac{2 \cdot 50.0}{(8.00)^2} = \frac{100.0}{64} = 1.5625 \, \text{m/s}^2
step 4
Now, using Newton's second law, we can relate the net force to the mass and acceleration: Fnet,x=ma=40.01.5625=62.5N F_{net,x} = m \cdot a = 40.0 \cdot 1.5625 = 62.5 \, \text{N} Setting the x-component equation equal to this value allows us to solve for F2 F_2 : 120cos(34)+F2cos(20)=62.5 120 \cos(34^\circ) + F_2 \cos(-20^\circ) = 62.5
step 5
After calculating F2 F_2 , we can find the coefficient of kinetic friction μk \mu_k using the equation: Ffriction=μkN F_{friction} = \mu_k \cdot N where N N is the normal force. The normal force can be found from the y-component equation. Rearranging gives us: μk=FfrictionN \mu_k = \frac{F_{friction}}{N}
Answer
The magnitude of force F2 F_2 is approximately 36.5 N, and the coefficient of kinetic friction is approximately 0.25.
Key Concept
Understanding the components of forces and applying Newton's laws is crucial in solving problems involving motion and friction.
Explanation
The solution involves breaking down the forces into components, calculating acceleration, and using Newton's second law to find unknown forces and coefficients of friction.
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