Question

A sled of mass 80 kg starts from rest and slides down an 18° incline 90 m long. It then travels for 20 m horizontally before starting back up a 16° incline. It travels 90 m along this incline before coming to rest. What is the net work done (in J) on the sled by friction?

Answer 1

Answer:

net work done by friction = - 2205 J

Explanation:

given,

mass of sled = 80 kg

slide down at an angle of 18°

length = 90 m

travel horizontally = 20 m before  starting back up a 16° incline.

net work = ?

$h_i = initial\ height$

$h_i = L_i Sin 18^0$  

$h_i = 90 sin 18^0$  

$h_i = 27.81\ m$

$h_f = Final\ height$

$h_f = L_f Sin 16^0$  

$h_f = 90 sin 16^0$  

$h_f =24.81\ m$

net work done by friction = change in potential energy

                                          = mg (h_f - h_i)

                                          = (75) (9.8) (- 27.81 + 24.81)  

net work done by friction = - 2205 J