Question

A box has a mass of 10.5 kg and has compressed a spring (with a spring constant of 71.7 N/m by 0.888m. The box is released from rest on a frictionless surface
a. How fast is the box moving when it leaves the spring?

b. After the box has left the spring it comes to a surface with
Mk = 0.234 How fare does it slide on this surface before stopping?

c. Instead of sliding to a stop it only goes 0.432 m on the rough surface and then returns to a frictionless surface. How fast is it moving?

Answer 1

Answer:

a. To find how fast it moves when it leaves the spring we use

1/2kx²= 1/2 mv²,

So making speed v subject

v=x√(k/m),

v= 0.888√(71.7/10.5) = 2.32 m/s

B) how far it goes after released

We know that

Work done = change in kinetic energy

So

(1/2 mv²0)= F x D,

F= mu x N,

But N=mg , F=0.234 x 10.5x 9.81= 24.10

So

0.5x 10.5 x 2.32 x 2.32= D x 24.10,

So

D= 1.17 m

C. Now using the work energy theorm:

0.5(mv1²- mv2²) =F xD = mu. mg x D

V2=√(V1²- 2gmu) =

√(2.32x 2.32-2 x 0.432 x 9.81 x 0.234) =

V2=1.8m/s