A car goes to the destination 180 km away from the starting place, drives 60 km at the original planned speed, and then drives at 1.5 times of the original speed. As a result, it reaches the destination 40 minutes earlier than the original plan, and calculates the original planned speed

A car goes to the destination 180 km away from the starting place, drives 60 km at the original planned speed, and then drives at 1.5 times of the original speed. As a result, it reaches the destination 40 minutes earlier than the original plan, and calculates the original planned speed

Suppose the original planned driving speed is x km / h, then: 180 − 60x − 180 − 601.5x = 4060, the solution is x = 60, and the test shows that x = 60 is the solution of the original equation and conforms to the meaning of the problem, so x = 60. Answer: the original planned driving speed is 60 km / h

A vehicle with a mass of 4T drives over a convex bridge deck with a radius of 50m. It is known that the vehicle always keeps a speed of 5m / s
When a vehicle with a mass of 4T passes a convex bridge deck with a radius of 50m, it always keeps a speed of 5m / s. The resistance of the vehicle is 0.05 times of the pressure of the vehicle on the bridge deck. What is the traction force of the vehicle when passing the highest point of the bridge n. (g = 10m / S2)

When it reaches the highest point, the pressure of the car on the bridge deck is equal to gravity (4 * 10 ^ 4N)
Because of the uniform speed, the resistance is equal to the traction force (4 * 10 ^ 4 * 0.05 = 200N), so the answer is 200N
The pressure of the car on the bridge deck changes continuously, and it is equal to gravity at the highest point

A vehicle with a mass of 4T passes a convex bridge with a radius of 50m at a constant speed
(1) What is the speed range of the car if it can pass the bridge safely?
(2) If the pressure on the bridge when the car passes through the highest point is equal to half of its gravity, what is the speed of the car at this time?
The detailed steps should be organized
There are answers

(1) When the speed is very high, the vehicle may have no pressure on the bridge due to weightlessness, which is very dangerous
mg=mvm^2/r
vm=√rg=√10*50=10√5=22.4m/s
If the vehicle can pass the bridge safely, its speed range is less than 22.4m/s
(2) If the pressure on the bridge when the car passes through the highest point is equal to half of its gravity, let the speed of the car be v
mg-mg/2=mv^2/r
v=√rg/2=√250=5√10=15.8m/s

When a vehicle with a mass of 4T passes a circular arch bridge with a radius of 50m at a speed of 5m / s, the dynamic friction coefficient between the bridge and the vehicle is μ = 0.5? （g=10m/s2）

According to Newton's second law: mg-N = mv2r, the solution support force is: n = mg − mv2r = 40000 − 4000 × 2550n = 38000n, then the friction force is: F = μ n = 0.5 × 38000n = 1.9 × 104n. A: when the car passes through the highest point of the bridge deck, the friction force is 1.9 × 104n