A mass m - 17 kg rests on a friction.ess table and acce.erated by a spring with

Question

A mass m - 17 kg rests on a friction.ess table and acce.erated by a spring with spnng constant k - 4277 N/m. The floor is fnctionless except for a rough patch. For this rcogh path, the coefficient of fnction is muk = 0.47. The mass leaves the spring at a peed v - 3.9 m/s. How much work is done by the spr-ng as it acce.erates the mass? How far was the spring stretched from its unst'eched length? The mass is measured to leave the rough spot with a fina. speed vf - 1.6 m/s. How much work is done by friction as the mass crosses the rough spot? What is the.ength of the rough spot? In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spnng comp'essed from its unstretched length? In this new scenano. what would the coefficient of fnction of the rough patch need to be changed to in order for the block to just barely make it through the rough patch? Return to a scenario where the blcok makes it through the entire rough patch. If the rough patch is lengthened to a distance of three times longer, as the block slides through the entire distance of the rough patch, the magnitude of the work dene by the force of friction is: the same three times greater three times less nine times greater nine times less

Explanation / Answer

1)

W = (1/2)mv^2 = (0.5) (17) (3.9)^2 = 129.285 J

2)

W = (1/2)kx^2

129.285 = (0.5) (4277) x^2

x = 0.2458 m

3)

W_f = (1/2)mv_i^2 - (1/2)mv_f^2

W_f = (0.5)(17)(3.9)^2 - (0.5)(17)(1.6)^2 = 129.285 - 21.76 = 107.525 J

4)

W_f = f_k d

W_f = _k Nd = _k mgd

107.525 = (0.5)(17)(9.8)d

d = 1.29 m

5)

Now (1/2)mv'_f^2 - (1/2)mv'_i^2 = - f_k(d/2)

(0.5)(15)(0) - (0.5)(17)v'_i^2 = - (107.525/2)

v'_i^2 = (107.525/2)/(8.5) = 6.325


(1/2)mv'_i^2 = (1/2)kx'^2

(17)(6.325) = 4277 x'^2

x' = 0.159 m

6)

(1/2)mv_i^2 = '_k mgd

(0.5)(17)(3.9^2) = '_k (17)(9.8)(1.29)

'_k = 0.60

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