A 2.0-kg cart and an 8.0-kg cart are connected by a relaxed, horizontal spring o

Question

A 2.0-kg cart and an 8.0-kg cart are connected by a relaxed, horizontal spring of spring constant300 N/m. You pull the 8.0-kg cart with some constant horizontal force. The separation between the carts increases for a short time interval, then remains constant as you continue to pull and the spring is stretched by 0.100 m.

A-What pulling force did you exert? Assume that the direction of your pull is the positive direction.

B-If you instead pull the 2.0-kg cart, what force must you exert to get the same stretch in the spring? Assume that the direction of your pull is the positive direction.

A is not 30 N

Explanation / Answer

as we are puling the 8 kg cart.
The force on the 2 kg cart is solely due to spring.

Now ,
as the spring is stretched by 0.1 m

so ,
force on 2 kg block = force due to spring = 300*.1 = 30 N

Now,

so , acceleration of the system = acc of 2 kg block = 30/2 = 15 m/s^2

Now,
acceleration of 8 kg block must also be equal to 15 ms^2

so

if we are applying F force ,
then,

F- F(due to spring) = 8* a

F - 30 = 8* 15

F = 150 N

2.
Now if we are pulling the 2 kg block,,
then , as spring force remains constant ,

acceleration of system = 30/8 = 3.75 m/s^2
( the force on the 8 kg cart is solely due to spring)

so ,

F - 30 = 2* 3.75

F = 37.5 N

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