A 14 kg box slides down a long, frictionless incline of angle30°. It starts from

Question

A 14 kg box slides down a long, frictionless incline of angle30°. It starts from rest at time t = 0 at the top ofthe incline at a height of 21 m above ground. (a) What is the original potential energy ofthe box relative to the ground?

(b) From Newton's laws, find the distance the box travels in 1 sand its speed at t = 1 s.

(c) Find the potential energy and the kinetic energy of the box att = 1 s.

(d) Find the kinetic energy and the speed of the box just as itreaches the bottom of the incline. Help would be appreciated!! Thankyou!!
(a) What is the original potential energy ofthe box relative to the ground?

(b) From Newton's laws, find the distance the box travels in 1 sand its speed at t = 1 s.

(c) Find the potential energy and the kinetic energy of the box att = 1 s.

(d) Find the kinetic energy and the speed of the box just as itreaches the bottom of the incline. Help would be appreciated!! Thankyou!!

Explanation / Answer

          Giventhat the mass of box is m = 14 kg           Heightof incline is h = 21 m           Angleof inclination is = 30o ------------------------------------------------------------- (a) The potential energy is U = mgh                                              =14 kg*9.8 m/s2*21m                                              =2881.2 J (b) But the accelaration of the body from Newtonlaw                                                         onthe incline plane is a = g sin                                                                                            =9.8m/s2*sin30o                                                                                             = 4.9 m/s2         From the equation of motion velocityafter time t =1s is V = u + at                                                                                             =0 +4.9 m/s2 *(1.0s)                                                                                             =4.9 m/s      The distance travelled is S =Ut + (1/2)at2                                                   =0 + (1/2)at2                                                    =(1/2)(4.9m/s2)(1.0s)2                                                      =---------- m (c) The kinetic energy at the t=1 is K =(1/2)mV2                                                              =------------ J      The potential energy at t =1sis P.E = Potential energy at height h - kinetic energy at t=1s                                                            = U - K                                                                  =----------- J (d) Apply conservation of energy at height h and bottomof the incline                               mgh = (1/2)mu2                                    u= 2gh                                       = --------- m/s                                    u= 2gh                                       = --------- m/s

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