A big frozen tofu turkey in a box of total mass m=5kg is pushed up an incline (f

Question

A big frozen tofu turkey in a box of total mass m=5kg is pushed up an incline (frictionless) a vertical distance of h=3m. . The block is traveling at the same speed (v1) at the bottom (1) of the incline as at the top (2). The speed at the top is v2=3m/s. At top the box slide a distance d1=15 cm on a frictionless surface until it encounters a friction patch of width d2=50 cm. Then the box slides again on a frictionless surface until it encounters a spring with spring constant k=500 N/m. The box compresses the spring a distance d3=10 cm before it comes momentarily to rest. Again all surfaces are frictionless except of the rough patch For all the questions in this problem use g=10m/s2 (a) How much work is done by the pushing force pushing the block up the incline? (10pt) (you don?t need the angle or Fpush ? use concepts you learned in ch 6 &7) (b) What is the total work done on the block by all forces on its route up the frictionless incline (1->2)? Explain (you don?t need the angle ? use concepts you learned in ch 6 &7) 10pt (c) Find the work done by friction over the length of the friction/rough patch. (state 2->3) (hint: look at the energy conservation from state (2) to state (3)) (25pt) (d) Find the coefficient of kinetic friction ?k in the rough patch. (10pt) (use Wf for work done by friction if you couldn?t find the answer in (c))

Explanation / Answer

a) work done = 0.5*m*v2^2+m*g*h =0.5*5*9+147=22.5+147=169.5J b)work is done by force of gravity which is mgh =147 J c)0.5*m*v2^2 =work done by friction + work done by spring work done by spring =0.5*k*x^2=2.5J =>work done by friction =22.5-2.5=20J d)20=Uk*mg*d2 [Uk is coefficient of kinetic friction] Uk =20/(5*9.8*.5)=.8163

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