Question

conservasion of energy

3. Toy Car Launcher A toy car has a mass of 0.75 kg. It is launched on a horizontal surface by using an elastic cord which is stretched 30 cm from its rest position. The spring constant of the cord is 1000 N/m. The car encounters a ramp and rolls up the ramp a distance x. The ramp angle is 20° relative to horizontal. Friction is negligible. (a) Which basic principle can be used in finding how far the car travels up the ramp? (b) How much energy is stored in the elastic cord immediately before the car is released? (c) Find the speed of the car while it is on the horizontal surface. (d) Find the impulse transmitted to the car by the spring. (e) Find the distance x which the car travels along the ramp 38

I have the answers but please show ALL work neatly. b) 45J c)10.95 m/s d)8.21 kg m/s e)17.9

Answer 1

Elastic energy of cord = potential energy of Block 1/2 k x^2_1 = mg sin theta x (a) Energy conservation (b) Energy steered = 1/2 times 1000 times . 3^2 = 45 joule (c) 1/2 m v^2 = 1/2 k x^2_1 v = squareroot k x^2_1/m = squareroot 1000 times . 3^2/.75 = squareroot 120 v = 10.95 m/s (d) I = m v = 0.75 times 10.95 I = 8.21 kg m/s (e) 1/2 k x^2 = mg sin theta x 1/2 times 1000 times 3^2 = 0.75 times 9.8 times sin 20 x x = 17.90 meter