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# PLEASE HELP?

A massless spring with force constant

604 N/m is fastened at its left end to a vertical

wall, as shown below.

The acceleration of gravity is 9.8 m/s

2

.

604 N/m

Initially, the 8 kg block and 5 kg block rest

on a horizontal surface with the 8 kg block in

contact with the spring (but not compressing it) and with the 5 kg block in contact with the

8 kg block. The 8 kg block is then moved to

the left, compressing the spring a distance of

0.6 m, and held in place while the 5 kg block

remains at rest as shown below.

2. The 8 kg block is then released and accelerates

to the right, toward the 5 kg block. The

surface is rough and the coefficient of friction

between each block and the surface is 0.5. The

two blocks collide, stick together, and move

to the right. Remember that the spring is not

attached to the 8 kg block.

Find the speed of the 8 kg block just before

it collides with the 5 kg block.

Answer in units of m/s.

3. Find the final speed of both blocks (stuck

together) just after they collide.

Answer in units of m/s.

4. Find the horizontal distance the blocks move

before coming to rest.

Answer in units of m.

### 1 Answer

- Anonymous1 month ago
2. Spring energy becomes KE and friction work:

½kx² = ½mv² + mgx

½ * 604N/m * (0.6m)² = ½ * 8kg * V² + 0.5*8kg*9.8m/s²*0.60

V = 4.6 m/s

3. conserve momentum:

8kg * 4.6m/s = (8+5)kg * v

v = 2.8 m/s

4. joint KE becomes friction work:

½Mv² = µMgd

½ * 13kg * (2.8m/s)² = 0.5 * 13kg * 9.8m/s² * d

and so

d = 0.82 m

Hope this helps!

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