A 0.27 kg toy car is held at rest against a 1796 N/m spring compressed a distance of 8cm. When released, the car travels a distance of 112cm along a flat surface before reaching a 0.30m high loop. The friction coefficient of the flat surface and the toy car is (0.4, 0.18). Assume friction is so small in the loop that it can be ignored.

Question: What is the velocity of the toy car when it reaches the top of the loop?

I know when it reaches the loop it is 6.91m/s, but I don't know what equation to use while it is in the loop?

Question

A 0.27 kg toy car is held at rest against a 1796 N/m spring compressed a distance of 8cm.  When released, the car travels a distance of 112cm along a flat surface before reaching a 0.30m high loop.  The friction coefficient of the flat surface and the toy car is (0.4, 0.18).  Assume friction is so small in the loop that it can be ignored.

Question:  What is the velocity of the toy car when it reaches the top of the loop?

I know when it reaches the loop it is 6.91m/s, but I don't know what equation to use while it is in the loop?

anonymous · Answer

Use law of conservation of energy.

anonymous · Answer

while going to the top of the loop the force of gravity acts on the car at 10m/s for 0.3 meters which induces a net force of 3 m/s... there fore it is traveling at 3.91m/s

anonymous · Answer

the total energy by the spring is $$kx ^{2}/2$$
where k is the spring constant
this energy is spent in doing work against friction firstly;
$$w = \mu _{k} \times resultant \times displacement$$
w is work
atop the hoop the total energy is
$$mg2r\ +\[mv ^{2}/2$$
equating all the equations we get the the velocity