PHYSICS MIDDLE SCHOOL

The spring is compressed a total of 3.0 cm, and used to set a 500 gram cart into motion. Find the speed of the cart at the instant it is released, assuming all the elastic potential energy is converted.

Answers

Answer 1

Answer:

Answer:

1.15 m/s

Explanation:

Part of the question is missing. Found the missing part on google:

"1. A hanging mass of 1500 grams compresses a spring 2.0 cm. Find the spring constant in N/m."

Solution:

First of all, we need to find the spring constant. We can use Hooke's law:

$F=kx$

where

$F=mg=(1.5 kg)(9.8 m/s^2)=14.7 N$ is the force applied to the spring (the weight of the hanging mass)

x = 2.0 cm = 0.02 m is the compression of the spring

Solving for k, we find the spring constant:

$k=(F)/(x)=(14.7)/(0.02)=735 N/m$

In the second part of the problem, the spring is compressed by

x = 3.0 cm = 0.03 m

So the elastic potential energy of the spring is

$U=(1)/(2)kx^2=(1)/(2)(735)(0.03)^2=0.33 J$

This energy is entirely converted into kinetic energy of the cart, which is:

$U=K=(1)/(2)mv^2$

where

m = 500 g = 0.5 kg is the mass of the cart

v is its speed

Solving for v,

$v=\sqrt{(2K)/(m)}=\sqrt{(2(0.33))/(0.5)}=1.15 m/s$

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Answers

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Answers

Answer:

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Answers

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Hope this helps!

How fast will a cart be moving if a force of 200 N is applied to a 50 kg cart for 10 sec?

Answers

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If there is no friction . . .

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Final speed = (speed of the cart before the force began) + (40 m/s more)

(That's about 89 mph faster than the cart was moving before the force started.)

I'm assuming that the cart starts from rest.

Initial velocity (vi): 0 m/s
Final velocity (vf): ?
Acceleration (a): ?
Force (F): 200 N
Mass (m): 50 kg
Time (t): 10 s

F= ma
200 N= (50kg)(a)
a= +4 m/s^2

a= (vf-vi)/t
+4 m/s^2= (vf - 0m/s)/10s
vf= 40 m/s

Answer: the cart will be moving 40 m/s.

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Answers

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