Asked by cole

A 65.0 kg clown gets into a 1250 kg clown car that has a 55.0 kg clown in the driver's seat. The car gets 1.40 cm closer to the ground.
A) What is the k value of the car's spring?
B) If two other clowns (45.0 kg and 52.0 kg) get into the car, how much closer with the car get to the ground?

Answers

Answered by bobpursley
k=force/distanc=(65+1250+55)*9.8/1.40 N/m

how much closer= (45+52)(9.8)/kabove
Answered by cole
the study guide says the answers are a) 4.5*10^4 N/m and b) 0.021 m but i still can't get those answers with your method
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