A very strange spring operates according to the graph shown here rather than obe

Question

A very strange spring operates according to the graph shown here rather than obeying Hooke?s Law. Note that the x-axis actually runs thru the middle of the graph (where F = 0). The axis labels are below the graph merely for clarity. The spring is employed as a launcher for a toy car. A child loads the car into the launcher, compressing the spring to x = ?5 cm.

a) When the trigger of the launcher is pulled, the spring accelerates the 50-g car horizontally to x= 0, where the car loses contact with the spring and exits the launcher. What work is done by that spring in accelerating the car? b) If the car subsequently moves freely (i.e. without friction) along a horizontal track, what final velocity will it have? c) The otherwise horizontal track contains a vertical loop of radius 15 cm. What will the car?s speed be at the top of the loop?

A very strange spring operates according to the graph shown here rather than obeying Hooke½s Law. Note that the x-axis actually runs thru the middle of the graph (where F = 0). The axis labels are below the graph merely for clarity. The spring is employed as a launcher for a toy car. A child loads the car into the launcher, compressing the spring to x = ?5 cm. a) When the trigger of the launcher is pulled, the spring accelerates the 50-g car horizontally to x= 0, where the car loses contact with the spring and exits the launcher. What work is done by that spring in accelerating the car? b) If the car subsequently moves freely (i.e. without friction) along a horizontal track, what final velocity will it have? c) The otherwise horizontal track contains a vertical loop of radius 15 cm. What will the car½s speed be at the top of the loop?

Explanation / Answer

Here ,

a)

Work done by spring = area under the curve

Work done by spring = 9*2*.01 + 0.5*20*0.01 + 0.5*0.02*8 + 0.5*0.02*2

Work done by spring = 0.38 J

the work done by spring is 0.38 J

b)

Here , Using Work energy theorum

0.5 * m*v^2 = Work done

0.5 * 0.050 * v^2 = 0.38

v = 3.9 m/s

the speed of car is 3.9 m/s

c)

Using third equation of motion ,

v^2 - u^2 = 2a s

v^2 - 3.9^2 = -2*9.8*0.30

v = 3.05 m/s

the speed of car at the top is 3.05 m/s

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