Question

8. A man pulls a 5.2-kg sled 49 meters along an angled hill with a force of 36 N, which elevates the man 31 meters above the bottom of the hill. The man then hops on his sled and slides from rest to the bottom of the hill back along his 49 meter path, during which a 301 N frictional force acts upon his sled. How much work in Joules does the man do against friction in pulling the sled up the hill?

Answer 1

7 K 49m mg 31 m do Free body diagram of bon, mgsino+f 36 N i fa frictional force mgroso Sino 31 49 0.6326 .. Coso . 0.5997 Netา mg sino t f = 36 5:2 x 9.81 X 0.6326 + f = 36 32.2702 36 n + → ats 3.7298 Work done by friction, Wf fxs = 3.72.98 X 4 9 Wy: 182.76 Work done by man against friction W = 182:76 J