3. A 230-kg piano rests on the ground. You want to get it into the back of a mov

Question

3. A 230-kg piano rests on the ground. You want to get it into the back of a moving-truck 1.0 meter above the ground. a. How much force would you (and your friends) need to exert on the piano if you lift it straight up through 1.0 m at constant speed? A. 230 N D. 1500 N B. 460 N E. 1900 N C. 720 N F. 2300 N b. How much force would you need to exert on the piano to push it up a frictionless 10.* ramp to a final vertical height of 1.0 m at constant speed? (Your force is exerted parallel to the ramp.) A. 18 N D. 140 N B. 26 N E. 230 N C. 84 N F. 390 N c. What is the mechanical advantage of using the ramp in part (b), compared to lifting the piano straight up? (Namely, what is the ratio of the two forces you calculated in parts (a)& (b)?) A. 3.2 D. 7.5 B. 4.0 E. 9.0 C. 5.8 F. 10. d. How much work would you (and your friends) do on the piano by lifting it straight up in part (a)? A. 1500 J D. 5300 J B. 2300 J E. 7800 C. 4400 J F. 8900 J e. How much work would you do on the piano by pushing it completely up the ramp in part (b)? A. 1500 J D. 5300 J B. 2300 J E. 7800 J C. 4400 J F. 8900 J Thought question [not for credit): Would you expect your answers to parts (d) & (e) to be equal or different? Ultimately, does having a large mechanical advantage- allow you to do less total work (hence, to expend less total energy) to perform a task? If not, then exactly what is the advantage?

Explanation / Answer

a) constant speed so fNet = 0

F - mg = 0

F = m g = 230*9.81=2256 N

so F)

b) now F - m g sin theta = 0

F = 2256*sin(10 degrees)=392 N

so F)

c)

2256/392=5.8

so C)

d) W = m g h = 2256*1= 2256 J

so B)

e)

W = F d = F h/sin theta = 392*1/sin(10 degrees)=2257 J

so B)

f) equal since you do the same work (sinc eyou get to same height) but you get to spend less force over a large distance

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