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?

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

if they have a finite resultant

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .