k=force/distanc=(65+1250+55)*9.8/1.40 N/m
how much closer= (45+52)(9.8)/kabove
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?
2 answers
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
1 answer