Question

A 1.0 kg puck sliding at 15 m/s along some horizontal frictionless ice strikes and compresses a horizontal spring attached to one end of the ice rink. If the spring has a constant of 35 N/m, what is the maximum compression of the spring?

Answer 1

\[\frac{ k }{ 2 }x^2=\frac{ 1 }{ 2 }mv_f^2-\frac{ 1 }{ 2 }mv_i^2\]
\[-35x^2=0-(1)15^2\]
find x

Answer 2

If in the above problem, the puck experiences a constant frictional force of 4.0 N opposing its motion beginning when it first strikes the spring, what would the maximum compression of the spring now be? (10 marks)

Answer 3

Aite man just chill a lil thanks

Answer 4

friction=4
work done by friction =4x
\[\frac{x^2k}{2}+4x=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2\]
\[-35x^2+8x=0-15^2\]

Answer 5

could you please explain the first one?

Answer 6

Y are you not dividing 35 by 2... and for 1/MV^2 u did not divide by 2

Answer 7

the force in the spring is thi given by \[\huge F_s=kx\]
and the work done by the spring force wich is its potential to move is given by
\[\huge W=\int\limits F_s=\int\limits \color{brown}{ kx} dx=\frac{1}{2}kx_f^2-\frac{1}{2}kx_i^2\]
so
we have
\[W_s=\Delta K\]
\[W_s=\frac{1}{2}kx_f^2-\frac{1}{2}kx_i^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2\]
x_f=0 v_f=0
so\[0-\frac{1}{2}kx_f^2=0-\frac{1}{2}mv_i^2\]
multiplying both sides by 2
\[kx^2=mv_i^2\]

Answer 8

@khally92