$W = \int{F(x)} \, dx = \int\limits^0_{-20} {(-kx)} \, dx = \frac{-kx^2}{2}\left \{ {{x=0} \atop {x=-20}} \right. = \frac{-60*(-20)^2}{2} \\W = -12000J$

Explanation:

Alternatively, the work done by the object is equal to the elastic potantial energy done by the spring.

$U = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 =0 - \frac{1}{2}60(-20)^2 = -12000J$

6 0

3 years ago

URGENT What does Newton's third law say about why momentum is conserved in collisions?

valentinak56 [21]

Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction.For such a collision<span>, the forces acting between the two objects are equal in magnitude and opposite in direction</span>

4 0

3 years ago

Read 2 more answers

Can someone please help me?

Luba_88 [7]

Answer:

1525 meters above ground

Explanation:

So to do this you will need to write this in slope intercept form or $y=mx+b$ . So 650 would be the b, 175 would be the m, and the x would be 5 so the equation would be $y=175(5)+650$ so if you solve or simplify the equation you will get 1525 meters above the ground and that would be our final answer.

5 0

3 years ago