Aspring has a force constant of 1 ã— 105 n/m. how far must it be stretched for its potential energy to be 47 j? answer in units of m.

= 0.03066 m

Work done or energy by a spring or an elastic material is given by the formula;

Potential energy = 1/2 kx²

Where, k is the spring constant and x is the distance stretched.

Therefore;

47 J = 1/2 (100000 n/m) x²

 47 J = 50, 000 x²

x² = 47 /50,000

x² = 0.00094

x = 0.03066 m

Therefore, x = 0.03066 m

given data:

determine the horse power (h.p) = ?

force applied to move the box(f) = 10 pounds,

box lifted from a distance         (d) = 3 feet

time taken to lift                         (t) = 10 sec,

the horse power is defined as amount of work required to move a load. the horse power is equal to 550 foot-pound per second

                  1 h.p   = 550 foot-pound/second

it can also be defined as the the load that need to be move a body for certain distance in a certain amount of time.

mathematically,

                      h.p = (force × distance)/ time

                            = (10 × 3)/ 10

                            = 30/10

                            = 3 foot -pound/sec

hence, the horse power is needed to move the box is 3 foot-pound/second

important note: 1 h.p = 550 foot-pound/second = 746 watts