Question

A block of mass m = 3.5 kg is on an inclined plane with a coefficient of friction μ₁ = 0.31, at an initial height h = 0.53 m above the ground. The plane is inclined at an angle θ = 54°. The block is then compressed against a spring a distance Δx = 0.11 m from its equilibrium point (the spring has a spring constant of k₁ = 39 N/m) and released. At the bottom of the inclined plane is a horizontal plane with a different coefficient of friction, μ₂ = 0.11. After a distance d = 0.19 m, there is another spring (spring constant k₂ = 6.5 N/m) under which the surface is frictionless. Use energy considerations in this problem.

a) Assume the block does not make it all the way to the bottom of the ramp before it comes to rest. What is the x-component of the block's position when it stops? Use an x-axis parallel to the incline, with the positive direction pointing down the incline and the origin at the spring's equilibrium position. This should not depend on the initial height of the block, only on the angle.

b) Let us now assume the values are such that the block does reach the bottom of the ramp. Find the speed of the block, in m/s, when it reaches the end of the ramp. Treat the block as a point.

c) Once the block reaches the bottom, it encounters a new coefficient of friction. What distance x, in meters, could the block travel on this surface before stopping if no obstacles were in the way? Assume the size of the block is negligible in transitioning from the ramp to the plane.

d) After a distance d on the surface with coefficient μ₂, the block encounters a frictionless surface and a spring with constant k₂. Calculate how far the spring is compressed Δs in m when the block comes to rest.

EXPEE ta (0%) Problem 13: A block of mass m - 3.5 kg is on an inclined plane with a coefficient of friction 11 = 0.31, at an initial height h=0.53 m above the ground. The plane is inclined at an angle 0= 54. The block is then compressed against a spring a distance Ar=0.11 m from its equilibrium point (the spring has a spring constant of k;= 39 Nim) and released. At the bottom of the inclined plane is a horizontal plane with a different coefficient of friction, 42 = 0.11. After a distance d= 0.19 m, there is another spring (spring constant k, = 6.5 N/m) under which the surface is frictionless. Use energy considerations in this problem. h Assignment Status A. Hy www Click here for detailed view d Otheexpertta.com 25% Part (a) Assume the block does not make it all the way to the bottom of the ramp before it comes to rest. What is the x-component of the block's position when it stops? Use an x-axis parallel to the incline, with the positive direction pointing down the incline and the origin at the spring's equilibrium position. This should not depend on the initial height of the block, only on the angle. Grade Summary Dechutions Problem Status 1 Completed 2 Completed 3 Completed 4 Completed 5 Completed 6 Completed 7 Completed 8 Completed 9 Completed 10 Completed 11 Completed 12 Completed 13 Partial 098 100% Pulenital 7 cosa) sina) COS(O) since) cos(0) sin(e) 8 9 HOME 4 5 6 2 3 + 0 END NORACKSPACE SPACE CLEAR d m Submissions Attempts Tenang 16 (0% per attempt) detailed view 09 2 09 026 8 t ki ? 4 Submit II Teedback I vive up! Hits: 0% deduction per hint. Hints remaining: - Feedback: 4 for a 0% deduction Your newer contains a sign error 25% Part (b) Let us now assume the values are such that the block does reach the bottom of the ramp. Find the speed of the block, in m/s, when it reaches the end of the ramp. Treat the block as a point. å 25% Part () Once the block reaches the bottom, it encounters a new coefficient of friction. What distance.x, in meters, could the block travel on this surface before stopping if no obstacles were in the way? Assume the size of the block is negligible in transitioning from the ramp to the plane. & 25% Part () After a distance d on the surface with coefficient M, the block encounters a frictionless surface and a spring with constant ks. Calculate how far the spring is compressed As in m when the block comes to rest. IMG-1433.jpg 1 IMG-1421.jpg Spring Constants...doc IMG-13002)jng 1 IMG-1894, ng Show all X

Answer 1

2 Sol. KA2923 -ngas sino + mg cos O-X a .. (= 0.5 KCAX22 Mimg coso - mg sino coordinate -C-02) Nicol)? AL 2 mg (ell Coso - sino - Δχ . Ki (D2C)? 2 mg cll, CUSO - Sino) 0.11 - 39(0.112 2×3.5×9.81 (0.31 Cos54 - sin54) 2 = 0,123 m Ang b. 1 K,Dx? = - mg (x+ox) sino + Mimg coso (+0) Ang 2 + 2 mua 0.5 x 3.5*V22 23.5 x 9.91% 0.233 sin 54. +0.5 x 39 x10.1002 70.31*3.5*9.84 Cos54 (0.233) i. Va 2.29 mis Ang 2.292 e / m v² = els mga => 2 = v2 Mag no 4.866 m a 1x3.5x (2.292 -0.11% 2.5x9.81 x 0.19 o.ll x g.81 2 = 2 6.5XDX? 2. 0x=1-61m Ans