A 1.24 kg block at rest on a tabletop is attached to a horizontal spring having
ID: 3279234 • Letter: A
Question
A 1.24 kg block at rest on a tabletop is attached to a horizontal spring having constant 19.2 N/m. The spring is initially unstretched. A constant 29.8 N horizontal force is applied to the object causing the spring to stretch. The acceleration of gravity is 9.8 m/s^2. Find the speed of the block after it has moved 0.238 m from equilibrium if the surface between block and tabletop is frictionless. Answer in units of m/s Find the speed of the block after it has moved 0.238 m from equilibrium if the coefficient of kinetic friction between block and tabletop is 0.104. Answer in units of m/s.Explanation / Answer
11) change in KE = total work done
= Fx - 0.5kx^2
= 29.8*0.238 - 0.5*19.2*0.238^2
= 6.5486
Final speed v = sqrt(2 KE/m) = sqrt (2*6.5486/1.24)
= 3.25 m/s answer
12) Here work done = Fx- 0.5kx^2 - umgx
= 29.8*0.238 - 0.5*19.2*0.238^2 - 0.104*1.24*9.8*0.238
= 6.2478 N
Speed v = sqrt (2 KE /m)
= sqrt (2*6.2478/1.24)
= 3.17 m/s answer
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