1. Consider a toy car track arrangement where a spring (k = 2000 N/m) launches a

Question

1. Consider a toy car track arrangement where a spring (k = 2000 N/m) launches a 100g toy car along a horizontal stretch of track and then the track curves by 90° to launch the car in a vertical direction. How far would the spring need to be compressed in order for the car to reach a height of 50 cm above the horizontal part of the track? (Ignore friction)

2. Consider a toy car track arrangement where a spring (k = 2000 N/m) launches the 100g toy car along a horizontal stretch of track and then the car enters a vertical circular loop of radius 15 cm. How far would the spring need to be compressed in order for the car to make it to the top of the loop and just remain in contact with the track at all times?

3. Now the car is launched onto a long stretch of horizontal track. The spring is compressed by 1 cm. What is the average friction force acting on the car if it comes to rest in 4m?

Explanation / Answer

Here,

spring constant , k = 2000 N/m

mass of car , m = 0.100 Kg

let the distance is x

height , h = 50 cm = 0.50 m

Using conservation of energy

0.5 *k * x^2 = m * g * h

0.5 * 2000 * x^2 = 0.10 * 9.8 * 0.50

x = 0.0211 m

the spring should be compressed by 0.0211 m

2)

spring constant , k = 2000 N/m

mass , m = 0.10 Kg

for the car to be at the at the track at the top

m * v^2/r = m * g

v^2/.15 = 9.8

v = 1.212 m/s

Using conservation of energy

0.5 * 2000 * x^2 = 0.5 * 0.1 * 1.212^2 + 0.1 * 9.8 * 0.15

x = 0.0148 m

the spring must be compressed by 0.0148 m

3)

let the average friction force is f

distance , d = 4 m

Using work energy

0.5 * k * x^2 = f * d

0.5 * 2000 * 0.01^2 = f * 4

f = 0.025 N

the frictional force is 0.025 N

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