Question

A 1.6 kg breadbox on a frictionless incline of angle ? = 33

Answer 1

----------------------------------------------------

The mass of the box m = 1.6kg

The spring constant k = 120N/m

The angle of the ramp ? = 33

(a) when the box slides 9cm

then the workdone W = mgsin?*x

                                  = (1.6)(9.8m/s^2)sin33*(0.09m)

                                 = 0.7686 J

now workdone is equal to

         1/2kx^2 + 1/2mv^2 = W

            (0.5)(120)(0.09)^2 + (0.5)(1.6)v^2 = 0.7686

             0.8v^2 = 0.2826

therefore the speed of the box

              v = 0.5943 m/s

(b) From law of conservation of energy

     1/2kx^2 + mgsin?*x = 0

   (0.5)(120)x = (1.6)(9.8)sin 33

Therefore the distance before its stopping

x = 0.142 m or 14.2cm

(c) From Newton's law

      the net force Fnet = T - mgsin?

and here T = kx   (spring force)

            ma =  kx -mgsin?

                 = (120)(0.142) - (1.6)(9.8)sin33

                 = 17.04-8.54 = 8.5

Therefore the acceleration a = 8.5/1.6 = 5.31 m/s^2

and its direction is upwards