A 425-g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 425-g block is dropped onto a relaxed vertical spring that has a spring constant k =120.0 N/m. The block becomes attached to the spring and compresses the spring 53.6 cm before momentarily stopping.

1. While the spring is being compressed, what work is done on the block by the gravitational force on it?

2. While the spring is being compressed, what work is done on the block by the spring force?

3. What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

4. If the speed at impact is doubled, what is the maximum compression of the spring?

Explanation / Answer

Here ,

spring constant , k = 120 N/m

mass of block , m = 0.425 Kg

distance , d = 0.536 m

1)

work dobe by gravitaional force = mg * h

work dobe by gravitaional force = 0.425 * 9.8 * 0.536

work dobe by gravitaional force = 2.23 J

the work dobe by gravitaional force is 2.23 J

2)

work done by spring force = -0.5 k x^2

work done by spring force = -0.5 * 120 * 0.536^2

work done by spring force = -17.24 J

the work done by spring force is -17.24 J

3)

let the speed if v

0.5 * 0.425 * v^2 = 17.24 - 2.23

v = 8.40 m/s

the speed of block is 8.4 m/s

4)

for double speed , maximum compression is x

0.425 * 9.8 * (x) + 0.5 * 0.425 * (2 * 8.4)^2 = 0.5 * 120 * x^2

solving for x

x = 1.035 m

the spring is compressed by 1.035 m

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