Question

7. A block of ice of mass m = 12 kg slides dom a frictionlons ramp at an angle 8 = 50° while an ice worker pulls on the block (via a rope) with & constant force that has a magnitude F 50N and is directed up the ramp. While this is happening the block slides down the incline through & distanced=0.50 m. Givens: m = 12 kg, F, -50N, d =0.50m, 8 = 50 g 9.8m/s? (a) Draw a Magnitude FBD for of the forces acting on the block. (b) De rive a symbolic expression for the work, W.(m .d.). for the work done on the block by the gravitational force. (e) Is the work done by the grav- itational force on the block is it slides positive, negative, or zero? Give a physical reason for your answer. (d) Derive a symbolic expression for the work, W. (m.g. d. 8. F), done on the block by the force that the rope exerts on the block. (e) Is the work done by the rope on the block as it slides positive, negative, or wro? Give a physical reason for your answer. (f) Derive a symbolic expression for the work. WNm, 3.d.). done on the block by the normal force (s) is the work done by the normal force on the block as it liden positive, negative, or xero? Give a physical reason for your answer. (h) Derive symbolic expression for the network, Wam,,d,F), done on the block as it slides through the displacement of magnitude (i) If the block starts from rest, use the work kinetic energy theorems to derive a symbolic expression for the speed, w....), of the block after it slides through the distance d ) Calculate a numerical value for the speed Check: If m 15kg F AN. 20.70m. 45 g = 9.8 ms.then G) 2.35m/s.

Answer 1

gr u sher mgcoso 16 wg mo.doo) = mgsino (4) ho = mg sino ft = ng caso So, Wg- masinad © wg (work done by gravita Honal foree) is posilve because mysino as pasaullel to the Displacement. Means force is acting along the direction of mohon

@ freson Wr(migi Oy Pa) = fred cos por 8 = 180° angle ble frand Wo 6m, 910, dif) = ford @ Negative because fr is en Hparallel to da